Đáp án:
2) x= \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
4)x= $\frac{5}{9}$
6)\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
8)\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
10)x=-1
Giải thích các bước giải:
2) $\frac{x+2}{x}$=$\frac{x^{2}+5x+4}{x^{2}+2x}$ +$\frac{x}{x+2}$
<=> $\frac{(x+2)^{2}}{x^{2}+2x}$- $\frac{x^{2}+5x+4}{x^{2}+2x}$- $\frac{x^{2}}{x^{2}+2x}$ =0
=> $x^{2}$+4x+4-$x^{2}$-5x-4- $x^{2}$ =0
<=>-$x^{2}$ -x=0
<=>-x(x+1)=0
=> x= \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
4)$\frac{3x-1}{x-1}$- $\frac{2x+5}{x-3}$ =1
<=> $\frac{(3x-1)(x-3)}{(x-1)(x-3)}$- $\frac{(2x+5)(x-1)}{(x-1)(x-3)}$ -$\frac{(x-1)(x-3)}{(x-1)(x-3)}$ =0 =>3$x^{2}$-10x+3-2$x^{2}$-3x+5-$x^{2}$+4x-3=0
=>-9x+5=0
=> x= $\frac{5}{9}$
6) $\frac{x}{2(x-3)}$+ $\frac{x}{2x+2}$= $\frac{2x}{(x+1)(x-3)}$
<=> $\frac{x(x+1)}{2(x-3)(x+1)}$+ $\frac{x(x-3)}{2(x-3)(x+1)}$-$\frac{4x}{2(x-3)(x+1)}$ =0 =>$x^{2}$+x+$x^{2}$-3x-4x=0
=>2$x^{2}$-6x=0
=>2x(x-3)=0
=>\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
8) $\frac{x+2}{x-2}$- $\frac{1}{x}$= $\frac{2}{x^{2}-2x}$
<=> $\frac{x(x+2)}{x^{2}-2x}$-$\frac{x-2}{x^{2}-2x}$- $\frac{2}{x^{2}-2x}$=0
=> $x^{2}$+2x -x+2-2=0
<=>$x^{2}$+x=0
<=>x(x+1)=0
=>\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
10) $\frac{2}{x-2}$+ $\frac{1}{(x+1)(x-2)}$= $\frac{x+2}{x^{2}-x-2}$
<=>$\frac{2(x+1)}{(x+1)(x-2)}$+ $\frac{1}{(x+1)(x-2)}$- $\frac{x+2}{(x+1)(x-2)}$=0
=>2x+2+1-x-2=0
=>x+1=0 =>x=-1
