Đề bài

Giải các phương trình sau: a) \(\left( {{x^2} - 4} \right) + x\left( {x - 2} \right) = 0;\) b) \({\left( {2x + 1} \right)^2} - 9{x^2} = 0.\)

Lời giải chi tiết

a) \(\left( {{x^2} - 4} \right) + x\left( {x - 2} \right) = 0;\) \(\begin{array}{l}\left( {{x^2} - 4} \right) + x\left( {x - 2} \right) = 0\\\left( {x - 2} \right)\left( {x + 2} \right) + x\left( {x - 2} \right) = 0\\\left( {x - 2} \right)\left( {x + 2 + x} \right) = 0\end{array}\) \(\begin{array}{l}\left( {x - 2} \right)\left( {2x + 2} \right) = 0\\TH1:x - 2 = 0\\x = 2\\TH2:2x + 2 = 0\\2x =  - 2\\x =  - 1\end{array}\) Vậy \(x \in \left\{ { - 1;2} \right\}.\) b) \({\left( {2x + 1} \right)^2} - 9{x^2} = 0.\) \(\begin{array}{l}{\left( {2x + 1} \right)^2} - {\left( {3x} \right)^2} = 0\\\left( {2x + 1 - 3x} \right)\left( {2x + 1 + 3x} \right) = 0\\(1-x).\left( {5x + 1} \right) = 0\end{array}\) \(TH1:1-x = 0\\x = 1\\TH2:5x + 1 = 0\\5x =- 1\\x =  -\frac{1}{5}\) Vậy \(x \in \left\{ { 1;-\frac{1}{5}} \right\}.\)