Bài 3 trang 5 sgk Toán lớp 8 tập 1

THanks

\(a)\ 3x(12x-4)-9x(4x-3)=30\)

\(\Leftrightarrow\) \(36x^2-12x-36x^2+27x=30\)

\(\Leftrightarrow\) \(15x=30\)

Vậy \(x=2.\)

\(b)\ x(5-2x)+2x(x-1)=15\)

\(\Leftrightarrow\) \(5x-2x^2+2x^2-2x=15\)

\(\Leftrightarrow\) \(3x=15\)

\(\Leftrightarrow\) \(x=5\)

Lời giải chi tiết

a. \(3x\left( {12x - 4} \right) - 9x\left( {4x - 3} \right) = 30\)

\(\begin{matrix} \Rightarrow 3x.12x - 3x.4 - 9x.4x + 9x.3 = 30 \hfill \\ \Rightarrow 36{x^2} - 12x - 36{x^2} + 27x = 30 \hfill \\ \Rightarrow \left( {36{x^2} - 36{x^2}} \right) + \left( { - 12x + 27x} \right) = 30 \hfill \\ \Rightarrow \left( {36 - 36} \right){x^2} + \left( { - 12 + 27} \right)x = 30 \hfill \\ \Rightarrow 0{x^2} + 15x = 30 \hfill \\ \Rightarrow 15x = 30 \hfill \\ \Rightarrow x = \dfrac{{30}}{{15}} = 2 \hfill \\ \end{matrix}\)

Vậy x = 2

b. \(x\left( {5 - 2x} \right) + 2x\left( {x - 1} \right) = 15\)

\(\begin{matrix} \Rightarrow x.5 - x.2x + 2x.x - 2x.1 = 15 \hfill \\ \Rightarrow 5x - 2{x^2} + 2{x^2} - 2x = 15 \hfill \\ \Rightarrow \left( { - 2{x^2} + 2{x^2}} \right) + \left( {5x - 2x} \right) = 15 \hfill \\ \Rightarrow \left( { - 2 + 2} \right){x^2} + \left( {5 - 2} \right)x = 15 \hfill \\ \Rightarrow 0{x^2} + 3x = 15 \hfill \\ \Rightarrow 3x = 15 \hfill \\ \Rightarrow x = \dfrac{{15}}{3} = 5 \hfill \\ \end{matrix}\)

Vậy x = 5