a: =>(x+6)(x-1)=0
=>x=-6 hoặc x=1
b: \(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
=>x=0(loại) hoặc x=-1(nhận)
a) \(x^2+5x-6=0\)
\(\left(x^2-x\right)+\left(6x-6\right)=0\)
\(\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đk:x\ne0;2\right)\)
\(\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(x^2+2x-x+2=2\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a, \(x^2+5x-6=0\Leftrightarrow x^2+6x-x-6=0\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
b, đk : x khác 0 ; 2
\(\Rightarrow x^2+2x-x+2=2\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0\left(ktm\right);x=-1\)
\(a,\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\\ b,Đk:x\ne0;-2\\ \Leftrightarrow x\left(x+2\right)-x+2=2\\ \Leftrightarrow x^2+x+2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
