a, x2+5x = 6
=> x2+5x - 6 =0
=> x2+6x -x- 6 =0
=> x(x+6)-(x+6)=0
=>(x-1)(x+6)=0
=> x=1 hoặc x=-6
b, x2-2015x +2014=0
=> x2-2014x-x +2014=0
=>x(x-2014)-(x-2014)=0
=> (x-1)(x-2014)=0
=> x=1 hoặc x=2014
k viết lại đề!
\(a.\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow x^2+6x-x-6=0\\ \Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\\ \Rightarrow S=\left\{1;-6\right\}\)
\(b.\\ \Leftrightarrow x^2-2014x-x+2014=0\\ \Leftrightarrow x\left(x-2014\right)-\left(x-2014\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2014\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2014=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2014\end{matrix}\right.\\ \Rightarrow S=\left\{1;2014\right\}\)
