Giải Toán 9 bài 4 Liên hệ giữa phép chia và phép khai phương

Giải SGK Toán 9 trang 18, 19, 20

Giải SGK Toán 9 bài 4: Liên hệ giữa phép chia và phép khai phương giúp các em trả lời các câu hỏi trong SGK Toán 9 tập 1 trang 18, 19, 20. Với hướng dẫn giải chi tiết giúp các em học sinh có thêm tài liệu tham khảo, so sánh đánh giá kết quả, từ đó học tốt môn Toán lớp 9. Dưới đây là chi tiết bài tập, các em tham khảo nhé.

Giải Toán 9 bài 4: Liên hệ giữa phép chia và phép khai phương

A. Trả lời câu hỏi trang 16, 17, 18 SGK Toán 9 tập 1
B. Giải bài tập trang 18, 19, 20 sgk Toán 9 tập 1
C. Trắc nghiệm Toán 9 bài 4

A. Trả lời câu hỏi trang 16, 17, 18 SGK Toán 9 tập 1

Câu hỏi 1 trang 16 SGK Toán 9 tập 1

Tính và so sánh: $\sqrt {\frac{{16}}{{25}}}$ $\sqrt {\frac{{16}}{{25}}}$ và $\frac{{\sqrt {16} }}{{\sqrt {25} }}$ $\frac{{\sqrt {16} }}{{\sqrt {25} }}$

Hướng dẫn giải:

$\left\{ {\begin{array}{*{20}{c}} {\sqrt {\dfrac{{16}}{{25}}} = \sqrt {\dfrac{{{4^2}}}{{{5^2}}}} = \sqrt {{{\left( {\dfrac{4}{5}} \right)}^2}} = \dfrac{4}{5}} \\ {\dfrac{{\sqrt {16} }}{{\sqrt {25} }} = \dfrac{{\sqrt {{4^2}} }}{{\sqrt {{5^2}} }} = \dfrac{4}{5}} \end{array}} \right. \Rightarrow \sqrt {\dfrac{{16}}{{25}}} = \dfrac{{\sqrt {16} }}{{\sqrt {25} }}$ $\left\{ {\begin{array}{*{20}{c}} {\sqrt {\dfrac{{16}}{{25}}} = \sqrt {\dfrac{{{4^2}}}{{{5^2}}}} = \sqrt {{{\left( {\dfrac{4}{5}} \right)}^2}} = \dfrac{4}{5}} \\ {\dfrac{{\sqrt {16} }}{{\sqrt {25} }} = \dfrac{{\sqrt {{4^2}} }}{{\sqrt {{5^2}} }} = \dfrac{4}{5}} \end{array}} \right. \Rightarrow \sqrt {\dfrac{{16}}{{25}}} = \dfrac{{\sqrt {16} }}{{\sqrt {25} }}$

Câu hỏi 2 trang 17 SGK Toán 9 tập 1

Tính:

a. $\sqrt {\frac{{225}}{{256}}}$ $\sqrt {\frac{{225}}{{256}}}$

b. $\sqrt {0,0196}$ $\sqrt {0,0196}$

Hướng dẫn giải:

a. $\sqrt {\frac{{225}}{{256}}} = \sqrt {\frac{{{{15}^2}}}{{{{16}^2}}}} = \sqrt {{{\left( {\frac{{15}}{{16}}} \right)}^2}} = \frac{{15}}{{16}}$ $\sqrt {\frac{{225}}{{256}}} = \sqrt {\frac{{{{15}^2}}}{{{{16}^2}}}} = \sqrt {{{\left( {\frac{{15}}{{16}}} \right)}^2}} = \frac{{15}}{{16}}$

b. $\sqrt {0,0196} = \sqrt {\frac{{196}}{{10000}}} = \sqrt {\frac{{{{14}^2}}}{{{{100}^2}}}} = \sqrt {{{\left( {\frac{{14}}{{100}}} \right)}^2}} = \frac{{14}}{{100}} = 0,14$ $\sqrt {0,0196} = \sqrt {\frac{{196}}{{10000}}} = \sqrt {\frac{{{{14}^2}}}{{{{100}^2}}}} = \sqrt {{{\left( {\frac{{14}}{{100}}} \right)}^2}} = \frac{{14}}{{100}} = 0,14$

Câu hỏi 3 trang 18 SGK Toán 9 tập 1

Tính:

a. $\frac{{\sqrt {999} }}{{\sqrt {111} }}$ $\frac{{\sqrt {999} }}{{\sqrt {111} }}$

b. $\frac{{\sqrt {52} }}{{\sqrt {117} }}$ $\frac{{\sqrt {52} }}{{\sqrt {117} }}$

Hướng dẫn giải:

a. $\frac{{\sqrt {999} }}{{\sqrt {111} }} = \sqrt {\frac{{999}}{{111}}} = \sqrt 9 = 3$ $\frac{{\sqrt {999} }}{{\sqrt {111} }} = \sqrt {\frac{{999}}{{111}}} = \sqrt 9 = 3$

b. $\frac{{\sqrt {52} }}{{\sqrt {117} }} = \sqrt {\frac{{52}}{{117}}} = \sqrt {\frac{4}{9}} = \frac{2}{3}$ $\frac{{\sqrt {52} }}{{\sqrt {117} }} = \sqrt {\frac{{52}}{{117}}} = \sqrt {\frac{4}{9}} = \frac{2}{3}$

Câu hỏi 3 trang 18 SGK Toán 9 tập 1

Rút gọn:

a. $\sqrt {\frac{{2{a^2}{b^4}}}{{50}}}$ $\sqrt {\frac{{2{a^2}{b^4}}}{{50}}}$ b. $\frac{{\sqrt {2a{b^2}} }}{{\sqrt {162} }}$ $\frac{{\sqrt {2a{b^2}} }}{{\sqrt {162} }}$ với $a \geqslant 0$ $a \geqslant 0$

Hướng dẫn giải:

a. $\sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \sqrt {\frac{{{a^2}{{\left( {{b^2}} \right)}^2}}}{{25}}} = \sqrt {{{\left( {\frac{{a{b^2}}}{5}} \right)}^2}} = \frac{{\left| a \right|.{b^2}}}{5}$ $\sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \sqrt {\frac{{{a^2}{{\left( {{b^2}} \right)}^2}}}{{25}}} = \sqrt {{{\left( {\frac{{a{b^2}}}{5}} \right)}^2}} = \frac{{\left| a \right|.{b^2}}}{5}$

Khi $a \geqslant 0 \Rightarrow \left| a \right| = a \Rightarrow \sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \frac{{a{b^2}}}{5}$ $a \geqslant 0 \Rightarrow \left| a \right| = a \Rightarrow \sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \frac{{a{b^2}}}{5}$

Khi $a < 0 \Rightarrow \left| a \right| = - a \Rightarrow \sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \frac{{ - a{b^2}}}{5}$ $a < 0 \Rightarrow \left| a \right| = - a \Rightarrow \sqrt {\frac{{2{a^2}{b^4}}}{{50}}} = \frac{{ - a{b^2}}}{5}$

B. Giải bài tập trang 18, 19, 20 sgk Toán 9 tập 1

Bài 28 trang 18 sgk Toán 9 tập 1

Tính

$a) \sqrt{\dfrac{289}{225}};$ $a) \sqrt{\dfrac{289}{225}};$ $b) \sqrt{2\dfrac{14}{25}};$ $b) \sqrt{2\dfrac{14}{25}};$

$c) \sqrt{\dfrac{0,25}{9}};$ $c) \sqrt{\dfrac{0,25}{9}};$ $d) \sqrt{\dfrac{8,1}{1,6}}.$ $d) \sqrt{\dfrac{8,1}{1,6}}.$

Hướng dẫn giải:

a) Ta có:

$\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{\sqrt {17^2}}{\sqrt{15^2}}=\dfrac{17}{15}.$ $\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{\sqrt {17^2}}{\sqrt{15^2}}=\dfrac{17}{15}.$

b) Ta có:

$\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{2.25+14}{25}}=\sqrt{\dfrac{50+14}{25}}$ $\sqrt{2\dfrac{14}{25}}=\sqrt{\dfrac{2.25+14}{25}}=\sqrt{\dfrac{50+14}{25}}$

$=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}.$ $=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}.$

c) Ta có:

$\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{\sqrt{0,5^2}}{\sqrt{3^2}}=\dfrac{0,5}{3}$ $\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{\sqrt{0,5^2}}{\sqrt{3^2}}=\dfrac{0,5}{3}$

$=0,5.\dfrac{1}{3}=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}.$ $=0,5.\dfrac{1}{3}=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}.$

d) Ta có:

$\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{81.0,1}{16.0,1}}=\sqrt{\dfrac{81}{16}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{\sqrt{9^2}}{\sqrt{4^2}}=\dfrac{9}{4}.$ $\sqrt{\dfrac{8,1}{1,6}}=\sqrt{\dfrac{81.0,1}{16.0,1}}=\sqrt{\dfrac{81}{16}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{\sqrt{9^2}}{\sqrt{4^2}}=\dfrac{9}{4}.$

Bài 29 trang 19 sgk Toán 9 tập 1

Tính

$a) \dfrac{\sqrt{2}}{\sqrt{18}};$ $a) \dfrac{\sqrt{2}}{\sqrt{18}};$ $b) \dfrac{\sqrt{15}}{\sqrt{735}};$ $b) \dfrac{\sqrt{15}}{\sqrt{735}};$

$c) \dfrac{\sqrt{12500}}{\sqrt{500}};$ $c) \dfrac{\sqrt{12500}}{\sqrt{500}};$ $d) \dfrac{\sqrt{6^{5}}}{\sqrt{2^{3}.3^{5}}}.$ $d) \dfrac{\sqrt{6^{5}}}{\sqrt{2^{3}.3^{5}}}.$

Hướng dẫn giải:

$a) \dfrac{\sqrt{2}}{\sqrt{18}}=\sqrt{\dfrac{2}{18}}=\sqrt{\dfrac{2.1}{2.9}}=\sqrt{\dfrac{1}{9}}=\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}} =\dfrac{1}{3}.$ $a) \dfrac{\sqrt{2}}{\sqrt{18}}=\sqrt{\dfrac{2}{18}}=\sqrt{\dfrac{2.1}{2.9}}=\sqrt{\dfrac{1}{9}}=\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}} =\dfrac{1}{3}.$

$b) \dfrac{\sqrt{15}}{\sqrt{735}}=\sqrt{\dfrac{15}{735}}=\sqrt{\dfrac{15.1}{15.49}}=\sqrt{\dfrac{1}{49}}=\sqrt {{{\left( {\dfrac{1}{7}} \right)}^2}}$ $b) \dfrac{\sqrt{15}}{\sqrt{735}}=\sqrt{\dfrac{15}{735}}=\sqrt{\dfrac{15.1}{15.49}}=\sqrt{\dfrac{1}{49}}=\sqrt {{{\left( {\dfrac{1}{7}} \right)}^2}}$

$=\dfrac{1}{7}.$ $=\dfrac{1}{7}.$

$c) \dfrac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\dfrac{12500}{500}}=\sqrt{\dfrac{500.25}{500}}$ $c) \dfrac{\sqrt{12500}}{\sqrt{500}}=\sqrt{\dfrac{12500}{500}}=\sqrt{\dfrac{500.25}{500}}$

$=\sqrt{25}=\sqrt{5^2}=5.$ $=\sqrt{25}=\sqrt{5^2}=5.$

$d) \dfrac{\sqrt{6^{5}}}{\sqrt{2^{3}.3^{5}}}=\sqrt{\dfrac{6^5}{2^3.3^5}}=\sqrt{\dfrac{(2.3)^5}{2^3.3^5}}=\sqrt{\dfrac{2^5.3^5}{2^3.3^5}}$ $d) \dfrac{\sqrt{6^{5}}}{\sqrt{2^{3}.3^{5}}}=\sqrt{\dfrac{6^5}{2^3.3^5}}=\sqrt{\dfrac{(2.3)^5}{2^3.3^5}}=\sqrt{\dfrac{2^5.3^5}{2^3.3^5}}$

$=\sqrt{\dfrac{2^5.3^5}{2^3.3^5}}=\sqrt{\dfrac{2^5}{2^3}}=\sqrt{\dfrac{2^3.2^2}{2^3}}=\sqrt{2^2}=2$ $=\sqrt{\dfrac{2^5.3^5}{2^3.3^5}}=\sqrt{\dfrac{2^5}{2^3}}=\sqrt{\dfrac{2^3.2^2}{2^3}}=\sqrt{2^2}=2$

Bài 30 trang 19 sgk Toán 9 tập 1

Rút gọn các biểu thức sau:

$a) \dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}$ $a) \dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}$ với $x > 0,\ y ≠ 0;$ $x > 0,\ y ≠ 0;$

$b) 2 y^{2}. \sqrt{\dfrac{x^{4}}{4y^{2}}}$ $b) 2 y^{2}. \sqrt{\dfrac{x^{4}}{4y^{2}}}$ với $y < 0;$

$c) 5xy. \sqrt{\dfrac{25x^{2}}{y^{6}}}$ $c) 5xy. \sqrt{\dfrac{25x^{2}}{y^{6}}}$ với $x < 0,\ y > 0;$ $x < 0,\ y > 0;$

$d) 0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}$ $d) 0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}$ với $x ≠ 0,\ y ≠ 0.$ $x ≠ 0,\ y ≠ 0.$

Hướng dẫn giải:

a) Ta có:

$\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^{4}}}$ $\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^{4}}}$

$=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{(y^2)^2}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}$ $=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{(y^2)^2}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}$

Vì $x> 0$ nên $|x|=x.$

Vì $y \ne 0$ $y \ne 0$ nên $y^2 > 0 \Rightarrow |y^2|=y^2.$ $y^2 > 0 \Rightarrow |y^2|=y^2.$

$\Rightarrow \dfrac{y}{x}.\dfrac{|x|}{|y^2|} =\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{y}{x}.\dfrac{x}{y.y}=\dfrac{1}{y}.$ $\Rightarrow \dfrac{y}{x}.\dfrac{|x|}{|y^2|} =\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{y}{x}.\dfrac{x}{y.y}=\dfrac{1}{y}.$

Vậy $\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{1}{y}.$ $\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{1}{y}.$

b) Ta có:

$2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{2^2.y^2}}$ $2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{2^2.y^2}}$

$=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{(2y)^2}}=2y^2.\dfrac{|x^2|}{|2y|}$ $=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{(2y)^2}}=2y^2.\dfrac{|x^2|}{|2y|}$

Vì $x^2 \ge 0 \Rightarrow |x^2|=x^2.$ $x^2 \ge 0 \Rightarrow |x^2|=x^2.$

Vì $y<0$ nên $2y < 0 \Rightarrow |2y|=-2y$ $2y < 0 \Rightarrow |2y|=-2y$

$\Rightarrow 2y^2.\dfrac{|x^2|}{|2y|}=2y^2.\dfrac{x^2}{-2y}=\dfrac{2y^2.x^2}{-2y}$ $\Rightarrow 2y^2.\dfrac{|x^2|}{|2y|}=2y^2.\dfrac{x^2}{-2y}=\dfrac{2y^2.x^2}{-2y}$

$=\dfrac{x^2.y.2y}{-2y}=-x^2y.$ $=\dfrac{x^2.y.2y}{-2y}=-x^2y.$

Vậy $2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=-x^2y.$ $2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=-x^2y.$

c) Ta có:

$5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=5xy.\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=\dfrac{\sqrt{5^2.x^2}}{\sqrt{(y^3)^2}}$ $5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=5xy.\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=\dfrac{\sqrt{5^2.x^2}}{\sqrt{(y^3)^2}}$

$=\dfrac{\sqrt{(5x)^2}}{\sqrt{(y^3)^2}}=5xy.\dfrac{|5x|}{|y^3|}$ $=\dfrac{\sqrt{(5x)^2}}{\sqrt{(y^3)^2}}=5xy.\dfrac{|5x|}{|y^3|}$

Vì $x<0 \Rightarrow |5x| = - 5x$ $x<0 \Rightarrow |5x| = - 5x$

Vì $y>0 \Rightarrow y^3 >0 \Rightarrow |y^3|=y^3.$ $y>0 \Rightarrow y^3 >0 \Rightarrow |y^3|=y^3.$

$\Rightarrow 5xy.\dfrac{|5x|}{|y^3|}=5xy.\dfrac{-5x}{y^3}=\dfrac{5xy.(-5x)}{y^3}$ $\Rightarrow 5xy.\dfrac{|5x|}{|y^3|}=5xy.\dfrac{-5x}{y^3}=\dfrac{5xy.(-5x)}{y^3}$

$=\dfrac{[5.(-5)].(x.x).y}{y^2.y}=\dfrac{-25x^2}{y^2}$ $=\dfrac{[5.(-5)].(x.x).y}{y^2.y}=\dfrac{-25x^2}{y^2}$

Vậy $5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=\dfrac{-25x^2}{y^2}.$ $5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=\dfrac{-25x^2}{y^2}.$

d) Ta có:

$0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}=0,2x^3y^3.\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}$ $0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}=0,2x^3y^3.\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}$

$=0,2x^3y^3\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2.(y^4)^2}}$ $=0,2x^3y^3\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2.(y^4)^2}}$

$=0,2x^3y^3.\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2}.\sqrt{(y^4)^2}}=0,2x^3y^3.\dfrac{4}{|x^2|.|y^4|}.$ $=0,2x^3y^3.\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2}.\sqrt{(y^4)^2}}=0,2x^3y^3.\dfrac{4}{|x^2|.|y^4|}.$

Bài 31 trang 19 sgk Toán 9 tập 1

Đề bài

a) So sánh $\sqrt{25 - 16}$ $\sqrt{25 - 16}$ và $\sqrt {25} - \sqrt {16};$ $\sqrt {25} - \sqrt {16};$

b) Chứng minh rằng: với $a > b >0$ thì $\sqrt a - \sqrt b < \sqrt {a - b} .$ $\sqrt a - \sqrt b < \sqrt {a - b} .$

Hướng dẫn giải:

a) Ta có:

$+) \sqrt {25 - 16} = \sqrt 9 =\sqrt{3^2}= 3.$ $+) \sqrt {25 - 16} = \sqrt 9 =\sqrt{3^2}= 3.$
$+) \sqrt {25} - \sqrt {16} = \sqrt{5^2}-\sqrt{4^2}=5 - 4 = 1 .$ $+) \sqrt {25} - \sqrt {16} = \sqrt{5^2}-\sqrt{4^2}=5 - 4 = 1 .$

Vì $3>1 \Leftrightarrow \sqrt {25 - 16}>\sqrt {25} - \sqrt {16} .$ $3>1 \Leftrightarrow \sqrt {25 - 16}>\sqrt {25} - \sqrt {16} .$

Vậy $\sqrt {25 - 16} > \sqrt {25} - \sqrt {16}$ $\sqrt {25 - 16} > \sqrt {25} - \sqrt {16}$

b Theo bài 26, ta đã chứng minh được: Với a>0 và b>0 thì:

$\sqrt{a+b}<\sqrt{a}+\sqrt{b}.$ $\sqrt{a+b}<\sqrt{a}+\sqrt{b}.$

Theo giải thiết, ta có

+) $b>0$

+) $a>b \Rightarrow a-b >0$ $a>b \Rightarrow a-b >0$

Áp dụng bài 26 cho hai số a-b và b, ta được:

$\sqrt{(a-b) +b}< \sqrt{a-b}+\sqrt{b}$ $\sqrt{(a-b) +b}< \sqrt{a-b}+\sqrt{b}$

$\Leftrightarrow \sqrt{a-b+b} < \sqrt{a-b} +\sqrt{b}$ $\Leftrightarrow \sqrt{a-b+b} < \sqrt{a-b} +\sqrt{b}$

$\Leftrightarrow \sqrt a < \sqrt{a-b}+\sqrt b$ $\Leftrightarrow \sqrt a < \sqrt{a-b}+\sqrt b$

$\Leftrightarrow \sqrt a - \sqrt b < \sqrt{a-b} (đpcm).$ $\Leftrightarrow \sqrt a - \sqrt b < \sqrt{a-b} (đpcm).$

Bài 32 trang 19 sgk Toán 9 tập 1

Tính

$a) \sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01};$ $a) \sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01};$

$b) \sqrt{1,44.1,21-1,44.0,4};$ $b) \sqrt{1,44.1,21-1,44.0,4};$

$c) \sqrt{\dfrac{165^{2}-124^{2}}{164}};$ $c) \sqrt{\dfrac{165^{2}-124^{2}}{164}};$

$d) \sqrt{\dfrac{149^{2}-76^{2}}{457^{2}-384^{2}}}.$ $d) \sqrt{\dfrac{149^{2}-76^{2}}{457^{2}-384^{2}}}.$

a) Ta có:

$\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{1.16+9}{16}.\dfrac{5.9+4}{9}.\dfrac{1}{100}}$ $\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{1.16+9}{16}.\dfrac{5.9+4}{9}.\dfrac{1}{100}}$

$=\sqrt{\dfrac{16+9}{16}.\dfrac{45+4}{9}.\dfrac{1}{100}}$ $=\sqrt{\dfrac{16+9}{16}.\dfrac{45+4}{9}.\dfrac{1}{100}}$

$=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.\dfrac{1}{100}}$ $=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.\dfrac{1}{100}}$

$=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}$ $=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}$

$=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}$ $=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}$

$=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}.\dfrac{\sqrt{7^2}}{\sqrt{3^2}}.\dfrac{1}{\sqrt{10^2}}$ $=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}.\dfrac{\sqrt{7^2}}{\sqrt{3^2}}.\dfrac{1}{\sqrt{10^2}}$

$=\dfrac{5}{4}.\dfrac{7}{3}.\dfrac{1}{10}=\dfrac{5.7.1}{4.3.10}=\dfrac{35}{120}=\dfrac{7}{24}.$ $=\dfrac{5}{4}.\dfrac{7}{3}.\dfrac{1}{10}=\dfrac{5.7.1}{4.3.10}=\dfrac{35}{120}=\dfrac{7}{24}.$

b) Ta có:

$\sqrt{1,44.1,21-1,44.0,4} = \sqrt{1,44(1,21-0,4)}$ $\sqrt{1,44.1,21-1,44.0,4} = \sqrt{1,44(1,21-0,4)}$

$=\sqrt{1,44.0,81}$ $=\sqrt{1,44.0,81}$

$=\sqrt{1,44}.\sqrt{0,81}$ $=\sqrt{1,44}.\sqrt{0,81}$

$=\sqrt{1,2^2}.\sqrt{0,9^2}$ $=\sqrt{1,2^2}.\sqrt{0,9^2}$

$=1,2.0,9=1,08.$

c) Ta có:

$\sqrt{\dfrac{165^{2}-124^{2}}{164}}=\sqrt{\dfrac{(165-124)(165+124)}{164}}$ $\sqrt{\dfrac{165^{2}-124^{2}}{164}}=\sqrt{\dfrac{(165-124)(165+124)}{164}}$

$=\sqrt{\dfrac{41.289}{41.4}}=\sqrt{\dfrac{289}{4}}$ $=\sqrt{\dfrac{41.289}{41.4}}=\sqrt{\dfrac{289}{4}}$

$=\dfrac{\sqrt{289}}{\sqrt{4}}=\dfrac{\sqrt{17^2}}{\sqrt{2^2}}=\dfrac{17}{2}.$ $=\dfrac{\sqrt{289}}{\sqrt{4}}=\dfrac{\sqrt{17^2}}{\sqrt{2^2}}=\dfrac{17}{2}.$

Câu d: Ta có:

$\sqrt{\dfrac{149^{2}-76^{2}}{457^{2}-384^{2}}}=\sqrt{\dfrac{(149-76)(149+76)}{(457-384)(457+384)}}$ $\sqrt{\dfrac{149^{2}-76^{2}}{457^{2}-384^{2}}}=\sqrt{\dfrac{(149-76)(149+76)}{(457-384)(457+384)}}$

$=\sqrt{\dfrac{73.225}{73.841}}=\sqrt{\dfrac{225}{841}}$ $=\sqrt{\dfrac{73.225}{73.841}}=\sqrt{\dfrac{225}{841}}$

Bài 33 trang 19 sgk Toán 9 tập 1

Giải phương trình

a) $\sqrt 2 .x - \sqrt {50} = 0;$ $\sqrt 2 .x - \sqrt {50} = 0;$

b) $\sqrt 3 .x + \sqrt 3 = \sqrt {12} + \sqrt {27};$ $\sqrt 3 .x + \sqrt 3 = \sqrt {12} + \sqrt {27};$

c) $\sqrt 3 .{x^2} - \sqrt {12} = 0;$ $\sqrt 3 .{x^2} - \sqrt {12} = 0;$

d) $\dfrac{x^2}{\sqrt 5 } - \sqrt {20} = 0$ $\dfrac{x^2}{\sqrt 5 } - \sqrt {20} = 0$

Hướng dẫn giải:

a)

$\sqrt{2}.x - \sqrt{50} = 0$ $\sqrt{2}.x - \sqrt{50} = 0$

$\Leftrightarrow \sqrt{2}x=\sqrt{50}$ $\Leftrightarrow \sqrt{2}x=\sqrt{50}$

$\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}$ $\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}$

$\Leftrightarrow x =\sqrt{\dfrac{50}{2}}$ $\Leftrightarrow x =\sqrt{\dfrac{50}{2}}$

$\Leftrightarrow x= \sqrt{25}$ $\Leftrightarrow x= \sqrt{25}$

$\Leftrightarrow x= \sqrt{5^2}$ $\Leftrightarrow x= \sqrt{5^2}$

$\Leftrightarrow x=5.$ $\Leftrightarrow x=5.$

Vậy $x=5.$

b)

$\sqrt{3}.x + \sqrt{3} = \sqrt{12} + \sqrt{27}$ $\sqrt{3}.x + \sqrt{3} = \sqrt{12} + \sqrt{27}$

$\Leftrightarrow \sqrt{3}.x = \sqrt{12} + \sqrt{27} - \sqrt{3}$ $\Leftrightarrow \sqrt{3}.x = \sqrt{12} + \sqrt{27} - \sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^3}. \sqrt{3}- \sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^3}. \sqrt{3}- \sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}$

$\Leftrightarrow \sqrt{3}.x=4\sqrt{3}$ $\Leftrightarrow \sqrt{3}.x=4\sqrt{3}$

$\Leftrightarrow x=4.$ $\Leftrightarrow x=4.$

Vậy $x=4.$

c)

$\sqrt{3}x^2-\sqrt{12}=0$ $\sqrt{3}x^2-\sqrt{12}=0$

$\Leftrightarrow \sqrt{3}x^2=\sqrt{12}$ $\Leftrightarrow \sqrt{3}x^2=\sqrt{12}$

$\Leftrightarrow \sqrt{3}x^2=\sqrt{4.3}$ $\Leftrightarrow \sqrt{3}x^2=\sqrt{4.3}$

$\Leftrightarrow \sqrt{3}x^2=\sqrt{4}.\sqrt 3$ $\Leftrightarrow \sqrt{3}x^2=\sqrt{4}.\sqrt 3$

$\Leftrightarrow x^2=\sqrt{4}$ $\Leftrightarrow x^2=\sqrt{4}$

$\Leftrightarrow x^2=\sqrt{2^2}$ $\Leftrightarrow x^2=\sqrt{2^2}$

$\Leftrightarrow x^2=2$ $\Leftrightarrow x^2=2$

$\Leftrightarrow \sqrt{x^2}=\sqrt{2}$ $\Leftrightarrow \sqrt{x^2}=\sqrt{2}$

$\Leftrightarrow |x|= \sqrt 2$ $\Leftrightarrow |x|= \sqrt 2$

$\Leftrightarrow x= \pm \sqrt 2.$ $\Leftrightarrow x= \pm \sqrt 2.$

Vậy $x= \pm\sqrt 2.$ $x= \pm\sqrt 2.$

d)

$\dfrac{x^{2}}{\sqrt{5}}- \sqrt{20} = 0$ $\dfrac{x^{2}}{\sqrt{5}}- \sqrt{20} = 0$

$\Leftrightarrow \dfrac{x^2}{\sqrt{5}}=\sqrt{20}$ $\Leftrightarrow \dfrac{x^2}{\sqrt{5}}=\sqrt{20}$

$\Leftrightarrow x^2=\sqrt{20}.\sqrt{5}$ $\Leftrightarrow x^2=\sqrt{20}.\sqrt{5}$

$\Leftrightarrow x^2=\sqrt{20.5}$ $\Leftrightarrow x^2=\sqrt{20.5}$

$\Leftrightarrow x^2=\sqrt{100}$ $\Leftrightarrow x^2=\sqrt{100}$

$\Leftrightarrow x^2=\sqrt{10^2}$ $\Leftrightarrow x^2=\sqrt{10^2}$

$\Leftrightarrow x^2=10$ $\Leftrightarrow x^2=10$

$\Leftrightarrow \sqrt{x^2}=\sqrt {10}$ $\Leftrightarrow \sqrt{x^2}=\sqrt {10}$

$\Leftrightarrow |x|=\sqrt{10}$ $\Leftrightarrow |x|=\sqrt{10}$

$\Leftrightarrow x=\pm \sqrt{10}.$ $\Leftrightarrow x=\pm \sqrt{10}.$

Vậy $x= \pm \sqrt{10}.$ $x= \pm \sqrt{10}.$

Bài 34 trang 19 sgk Toán 9 tập 1

Rút gọn các biểu thức sau:

a) $ab^{2}.\sqrt{\dfrac{3}{a^{2}b^{4}}}$ $ab^{2}.\sqrt{\dfrac{3}{a^{2}b^{4}}}$ với $a < 0,\ b ≠ 0;$ $a < 0,\ b ≠ 0;$

b) $\sqrt{\dfrac{27(a - 3)^{2}}{48}}$ $\sqrt{\dfrac{27(a - 3)^{2}}{48}}$với $a > 3;$

c) $\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}$ $\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}$ với $a ≥ -1,5$ và $b < 0.$

d) $(a - b).\sqrt{\dfrac{ab}{(a - b)^{2}}}$ $(a - b).\sqrt{\dfrac{ab}{(a - b)^{2}}}$ với $a < b < 0.$

Hướng dẫn giải:

a) Ta có:

$(ab^{2}.\sqrt{\dfrac{3}{a^{2}b^{4}}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}$ $(ab^{2}.\sqrt{\dfrac{3}{a^{2}b^{4}}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}$

$=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{(b^2)^2}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}$ $=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{(b^2)^2}}=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}$

$=ab^2.\dfrac{\sqrt{3}}{-ab^2}=-\sqrt{3}.$ $=ab^2.\dfrac{\sqrt{3}}{-ab^2}=-\sqrt{3}.$

(Vì $a < 0$ nên $|a|=-a$ và $b \ne 0$ $b \ne 0$ nên $b^2 >0 \Rightarrow |b^2|=b^2) .$ $b^2 >0 \Rightarrow |b^2|=b^2) .$

b) Ta có:

$\sqrt{\dfrac{27(a - 3)^{2}}{48}}=\sqrt{\dfrac{27}{48}.(a-3)^2}=\sqrt{\dfrac{27}{48}}.\sqrt{(a-3)^2}$ $\sqrt{\dfrac{27(a - 3)^{2}}{48}}=\sqrt{\dfrac{27}{48}.(a-3)^2}=\sqrt{\dfrac{27}{48}}.\sqrt{(a-3)^2}$

$=\sqrt{\dfrac{9.3}{16.3}}.\sqrt{(a-3)^2}=\sqrt{\dfrac{9}{16}}.\sqrt{(a-3)^2}$ $=\sqrt{\dfrac{9.3}{16.3}}.\sqrt{(a-3)^2}=\sqrt{\dfrac{9}{16}}.\sqrt{(a-3)^2}$

$=\sqrt{\dfrac{3^2}{4^2}}.\sqrt{(a-3)^2}=\dfrac{\sqrt {3^2}}{\sqrt {4^2}}.\sqrt{(a-3)^2}$ $=\sqrt{\dfrac{3^2}{4^2}}.\sqrt{(a-3)^2}=\dfrac{\sqrt {3^2}}{\sqrt {4^2}}.\sqrt{(a-3)^2}$

$=\dfrac{3}{4}|a-3|=\dfrac{3}{4}(a-3).$ $=\dfrac{3}{4}|a-3|=\dfrac{3}{4}(a-3).$

( Vì $a > 3$ nên $a-3>0 \Rightarrow |a-3|=a-3)$ $a-3>0 \Rightarrow |a-3|=a-3)$

c) Ta có:

$\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=\sqrt{\dfrac{3^2+2.3.2a+2^2.a^2}{b^2}}$ $\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=\sqrt{\dfrac{3^2+2.3.2a+2^2.a^2}{b^2}}$

$=\sqrt{\dfrac{3^2+2.3.2a+(2a)^2}{b^2}}=\sqrt{\dfrac{(3+2a)^2}{b^2}}$ $=\sqrt{\dfrac{3^2+2.3.2a+(2a)^2}{b^2}}=\sqrt{\dfrac{(3+2a)^2}{b^2}}$

$=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}$ $=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}$

Vì $a \geq -1,5 \Rightarrow a+1,5>0$ $a \geq -1,5 \Rightarrow a+1,5>0$

$\Leftrightarrow 2(a+1,5)>0$ $\Leftrightarrow 2(a+1,5)>0$

$\Leftrightarrow 2a+3>0$ $\Leftrightarrow 2a+3>0$

$\Leftrightarrow 3+2a>0$ $\Leftrightarrow 3+2a>0$

$\Rightarrow |3+2a|=3+2a$ $\Rightarrow |3+2a|=3+2a$

Vì $b<0\Rightarrow |b|=-b$ $b<0\Rightarrow |b|=-b$

Do đó: $\dfrac{|3+2a|}{|b|}=\dfrac{3+2a}{-b} =-\dfrac{3+2a}{b}.$ $\dfrac{|3+2a|}{|b|}=\dfrac{3+2a}{-b} =-\dfrac{3+2a}{b}.$

Vậy $\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=-\dfrac{3+2a}{b}.$ $\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=-\dfrac{3+2a}{b}.$

d) Ta có:

$(a - b).\sqrt{\dfrac{ab}{(a - b)^{2}}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}$ $(a - b).\sqrt{\dfrac{ab}{(a - b)^{2}}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}$

$=(a-b).\dfrac{\sqrt{ab}}{|a-b|}$ $=(a-b).\dfrac{\sqrt{ab}}{|a-b|}$

$=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=-\sqrt{ab}.$ $=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=-\sqrt{ab}.$

Bài 35 trang 20 sgk Toán 9 tập 1

Tìm x, biết:

a) $\sqrt {{{\left( {x - 3} \right)}^2}} = 9;$ $\sqrt {{{\left( {x - 3} \right)}^2}} = 9;$

b) $\sqrt {4{{\rm{x}}^2} + 4{\rm{x}} + 1} = 6.$ $\sqrt {4{{\rm{x}}^2} + 4{\rm{x}} + 1} = 6.$

Hướng dẫn giải:

a) Ta có:

$\sqrt {{{\left( {x - 3} \right)}^2}} = 9 \Leftrightarrow \left| {x - 3} \right| = 9$ $\sqrt {{{\left( {x - 3} \right)}^2}} = 9 \Leftrightarrow \left| {x - 3} \right| = 9$

$\Leftrightarrow \left[ \matrix{ x - 3 = 9 \hfill \cr x - 3 = - 9 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 9 + 3 \hfill \cr x = - 9 + 3 \hfill \cr} \right.$ $\Leftrightarrow \left[ \matrix{ x - 3 = 9 \hfill \cr x - 3 = - 9 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 9 + 3 \hfill \cr x = - 9 + 3 \hfill \cr} \right.$

$\Leftrightarrow \left[ \matrix{ x = 12 \hfill \cr x = - 6 \hfill \cr} \right.$ $\Leftrightarrow \left[ \matrix{ x = 12 \hfill \cr x = - 6 \hfill \cr} \right.$

Vậy phương trình đã cho có hai nghiệm: $x = 12$ và $x = -6.$

b) Ta có:

$\sqrt{4x^2+4x+1}=6 \Leftrightarrow \sqrt{2^2x^2+4x+1}=6$ $\sqrt{4x^2+4x+1}=6 \Leftrightarrow \sqrt{2^2x^2+4x+1}=6$

$\Leftrightarrow \sqrt{(2x)^2+2.2x+1^2}=6$ $\Leftrightarrow \sqrt{(2x)^2+2.2x+1^2}=6$

$\Leftrightarrow \sqrt{(2x+1)^2}=6$ $\Leftrightarrow \sqrt{(2x+1)^2}=6$

$\Leftrightarrow |2x+1| =6$ $\Leftrightarrow |2x+1| =6$

$\eqalign{ & \Leftrightarrow \left[ \matrix{ 2x + 1 = 6 \hfill \cr 2x + 1 = - 6 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 2x = 6 - 1 \hfill \cr 2x = - 6 - 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 2x = 5 \hfill \cr 2x = - 7 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = \dfrac{5}{2} \hfill \cr x = \dfrac{-7}{2} \hfill \cr} \right. \cr} .$ $\eqalign{ & \Leftrightarrow \left[ \matrix{ 2x + 1 = 6 \hfill \cr 2x + 1 = - 6 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 2x = 6 - 1 \hfill \cr 2x = - 6 - 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 2x = 5 \hfill \cr 2x = - 7 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = \dfrac{5}{2} \hfill \cr x = \dfrac{-7}{2} \hfill \cr} \right. \cr} .$

Vậy phương trình có 2 nghiệm $x = \dfrac{5}{2}$ $x = \dfrac{5}{2}$ và $x=\dfrac{-7}{2}.$ $x=\dfrac{-7}{2}.$

Bài 36 trang 20 SGK Toán 9 tập 1

Mỗi khẳng định sau đúng hay sai? Vì sao?

$a) 0,01 = \sqrt {0,0001} ;$ $a) 0,01 = \sqrt {0,0001} ;$

$b) - 0,5 = \sqrt { - 0,25} ;$ $b) - 0,5 = \sqrt { - 0,25} ;$

$c) \sqrt {39} < 7$ $c) \sqrt {39} < 7$ và $\sqrt {39} > 6;$ $\sqrt {39} > 6;$

$d) \left( {4 - 13} \right).2{\rm{x}} < \sqrt 3 \left( {4 - \sqrt {13} } \right) \Leftrightarrow 2{\rm{x}} < \sqrt {3} .$ $d) \left( {4 - 13} \right).2{\rm{x}} < \sqrt 3 \left( {4 - \sqrt {13} } \right) \Leftrightarrow 2{\rm{x}} < \sqrt {3} .$

Hướng dẫn giải:

a) Đúng.

Vì $VP=\sqrt{0,0001}=\sqrt{0,01^2}=0,01=VT.$ $VP=\sqrt{0,0001}=\sqrt{0,01^2}=0,01=VT.$

b) Sai.

Vì số âm không có căn bậc hai.

c) Đúng. Vì:

$\left\{ \matrix{ {6^2} = 36 \hfill \cr {\left( {\sqrt {39} } \right)^2} = 39 \hfill \cr {7^2} = 49 \hfill \cr} \right.$ $\left\{ \matrix{ {6^2} = 36 \hfill \cr {\left( {\sqrt {39} } \right)^2} = 39 \hfill \cr {7^2} = 49 \hfill \cr} \right.$

Mà $36 < 39 < 49\Leftrightarrow \sqrt {36} < \sqrt {39} < \sqrt {49}$ $36 < 39 < 49\Leftrightarrow \sqrt {36} < \sqrt {39} < \sqrt {49}$

$\Leftrightarrow \sqrt {{6^2}} < \sqrt {39} < \sqrt {{7^2}}$ $\Leftrightarrow \sqrt {{6^2}} < \sqrt {39} < \sqrt {{7^2}}$

$\Leftrightarrow 6 < \sqrt {39} < 7$ $\Leftrightarrow 6 < \sqrt {39} < 7$

Hay $\sqrt{39}>6$ $\sqrt{39}>6$ và $\sqrt{39} < 7.$ $\sqrt{39} < 7.$

d) Đúng.

Xét bất phương trình đề cho:

$(4-\sqrt{13}).2x<\sqrt 3 .(4-\sqrt{13})(1)$ $(4-\sqrt{13}).2x<\sqrt 3 .(4-\sqrt{13})(1)$

Ta có:

$\left\{ \matrix{ {4^2} = 16 \hfill \cr {\left( {\sqrt {13} } \right)^2} = 13 \hfill \cr} \right.$ $\left\{ \matrix{ {4^2} = 16 \hfill \cr {\left( {\sqrt {13} } \right)^2} = 13 \hfill \cr} \right.$

Mà $16>13 \Leftrightarrow \sqrt{16} > \sqrt{13}$ $16>13 \Leftrightarrow \sqrt{16} > \sqrt{13}$

$\Leftrightarrow \sqrt{4^2}> \sqrt{13}$ $\Leftrightarrow \sqrt{4^2}> \sqrt{13}$

$\Leftrightarrow 4> \sqrt{13}$ $\Leftrightarrow 4> \sqrt{13}$

$\Leftrightarrow 4-\sqrt{13}>0$ $\Leftrightarrow 4-\sqrt{13}>0$

Chia cả hai vế của bất đẳng thức $(1)$ cho số dương $(4-\sqrt{13})$ $(4-\sqrt{13})$, ta được:

$\dfrac{(4-\sqrt{13}).2x}{(4-\sqrt{13})} <\dfrac{\sqrt 3 .(4-\sqrt{13})}{(4-\sqrt{13})}$ $\dfrac{(4-\sqrt{13}).2x}{(4-\sqrt{13})} <\dfrac{\sqrt 3 .(4-\sqrt{13})}{(4-\sqrt{13})}$

$\Leftrightarrow 2x < \sqrt 3.$ $\Leftrightarrow 2x < \sqrt 3.$

Vậy phép biến đổi tương đương trong câu d là đúng.