Tìm x, y, biết: a) x^2 +7x = 8 b) 2x^2 +4y^2 +10x + 4xy = -25

* Hướng dẫn giải

a)

$\begin{array}{l}{x}^2+7x=8\\ x^2+7x-8=0\\ x^2+8x-x-8=0\\ x\left(x+8\right)-\left(x+8\right)=0\\ \left(x-1\right)\left(x+8\right)=0\\ \Rightarrow \left[\begin{array}{l}x-1=0\\ x+8=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=1\\ x=-8\end{array}\right.\end{array}$

Vậy $x\in \left\{-8;1\right\}$

b)

$\begin{array}{l}2x^2+4y^2+10x+4xy=-25\\ 2x^2+4y^2+10x+4xy+25=0\\ \left(x^2+10x+25\right)+\left(x^2+4xy+4y^2\right)=0\\ {\left(x+5\right)}^2+{\left(x+2y\right)}^2=0\end{array}$