Bài 8 trang 10 sgk Toán lớp 9 tập 1

a. \(\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right|\)

Do \(2 - \sqrt 3 > 0 \Rightarrow \left| {2 - \sqrt 3 } \right| = 2 - \sqrt 3\)

\(\Rightarrow \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = 2 - \sqrt 3\)

b. \(\sqrt {{{\left( {3 - \sqrt {11} } \right)}^2}} = \left| {3 - \sqrt {11} } \right|\)

Do \(3 - \sqrt {11} < 0 \Rightarrow \left| {3 - \sqrt {11} } \right| = \sqrt {11} - 3\)

\(\Rightarrow \sqrt {{{\left( {3 - \sqrt {11} } \right)}^2}} = \sqrt {11} - 3\)

c. \(2\sqrt {{a^2}} = 2\left| a \right|\)

Do \(a \geqslant 0 \Rightarrow \left| a \right| = a\)

\(\Rightarrow 2\sqrt {{a^2}} = 2a\)

d. \(3\sqrt {{{\left( {a - 2} \right)}^2}} = 3\left| {a - 2} \right|\)

Do \(a < 2 \Rightarrow a - 2 < 0 \Rightarrow \left| {a - 2} \right| = 2 - a\)

\(\Rightarrow 3\sqrt {{{\left( {a - 2} \right)}^2}} = 3\left( {2 - a} \right) = 6 - 3a\)

a)

vì \(\left\{ \matrix{{2^2} = 4 \hfill \cr {\left( {\sqrt 3 } \right)^2} = 3 \hfill \cr} \right.\)

mà \(4>3\) nên \(\sqrt{4} > \sqrt{3} \Leftrightarrow 2> \sqrt{3} \Leftrightarrow 2- \sqrt{3}>0.\)

\(\Leftrightarrow \left| {2 - \sqrt 3 } \right| =2- \sqrt{3}.\)

Do đó: \(\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right|=2- \sqrt{3}\)

b)

Vì \(\left\{ \matrix{{3^2} = 9 \hfill \cr {\left( {\sqrt {11} } \right)^2} = 11 \hfill \cr} \right.\)

mà \(9<11\) nên \(\sqrt{9} < \sqrt{11} \Leftrightarrow 3< \sqrt{11} \Leftrightarrow 3- \sqrt{11} <0\)

\(\Leftrightarrow \left| {3 - \sqrt {11} } \right| =-(3- \sqrt{11})=-3+\sqrt{11}\)

\(=\sqrt{11}-3.\)

Do đó: \(\sqrt {{{\left( {3 - \sqrt {11} } \right)}^2}} = \left| {3 - \sqrt {11} } \right| =\sqrt{11}-3.\)

c) Ta có: \(2\sqrt {{a^2}} = 2\left| a \right| = 2{\rm{a}}\) (vì \(a \ge 0\) )

d) Vì \(a < 2\) nên \(a - 2<0\).

\(\Leftrightarrow \left| a-2 \right|=-(a-2)=-a+2=2-a\)

Do đó: \(3\sqrt {{{\left( {a - 2} \right)}^2}} = 3\left| {a - 2} \right| = 3\left( {2 - a} \right)\)

\(= 6 - 3a.\)

Thank nha