Tìm x, y, biết: a) x^2 +4x = 0 b) 5x (3x -2) = 4 -9x^2

* Hướng dẫn giải

a)

$\begin{array}{l}{x}^2+4x=0\\ x\left(x+4\right)=0\\ \Rightarrow \left[\begin{array}{l}x=0\\ x+4=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=0\\ x=-4\end{array}\right.\end{array}$

Vậy $x\in \left\{-4;0\right\}$

b)

$\begin{array}{l}5x\left(3x-2\right)=4-9x^2\\ 5x\left(3x-2\right)=\left(2-3x\right)\left(2+3x\right)\\ 5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\\ 5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\\ \left(3x-2\right)\left(5x+2+5x\right)=0\\ \left(3x-2\right)\left(8x+2\right)=0\\ \Rightarrow \left[\begin{array}{l}3x-2=0\\ 8x+2=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=\frac{2}{3}\\ x=\frac{-1}{4}\end{array}\right.\end{array}$

Vậy $x\in \left\{\frac{-1}{4};\frac{2}{3}\right\}$