Question

[Tuyết Ly]

Giải phương trình sau:

a) x²+5x-6=0

b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

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)

Answer 2

[Minh Hiếu]

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) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đk:x\ne0;2\right)\)

\(\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

\(x^2+2x-x+2=2\)

\(x^2+x=0\)

\(x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)