Đáp án:
Giải thích các bước giải:
a) ⇔( 3x-1)(x-2)-(x-2)(x+1)=0
⇔(3x-1-x-1)(x-2)=0
⇔2x(x-2)=0
⇔\(\left[ \begin{array}{l}2x=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy ......
b, ĐKXĐ: x$\neq$ -1, x$\neq$ 2
⇔$\frac{3(x-2)}{(x+1)(x-2)}$ - $\frac{2(x+1)}{(x+1)(x-2)}$ =$\frac{4x-2}{(x+1)(x-2)}$
⇒3x-6-2x-2=4x-2
⇔x-8=4x-2
⇔x-4x=-2+8
⇔-3x=6
⇔x=-2
Vậy.......
c) ĐKXĐ x≥$\frac{-5}{2}$
⇔\(\left[ \begin{array}{l}x-9=2x+5\\x-9=-2x-5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x-2x=5+9\\x+2x=-5+9\end{array} \right.\)
⇔\(\left[ \begin{array}{l}-x=14\\3x=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-14(ko thỏa mãn)\\x=\frac{4}{3} (thỏa mãn)\end{array} \right.\)
Vậy
