Đáp án:
Giải thích các bước giải:
ta có:
a,$x^{2}$-5x-6=0
⇔$x^{2}$+x-6x-6=0
⇔x(x+1)-6(x+1)=0
⇔(x+1)(x-6)=0
⇔\(\left[ \begin{array}{l}x+1=0\\x-6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=6\end{array} \right.\)
b,ĐKXĐ:x$\neq$0,2
$\frac{x+2}{x-2}$-$\frac{1}{x}$=$\frac{2}{x(x-2)}$
⇔$\frac{x(x+2)}{x(x-2)}$-$\frac{1(x-2)}{x(x-2)}$=$\frac{2}{x(x-2)}$
⇔x(x+2)-1(x-2)=2
⇔$x^{2}$+2x-x+2=2
⇔$x^{2}$+x=0
⇔x(x+1)=0
⇔\(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0)ktm)\\x=-1(tm)\end{array} \right.\)
