\(a,\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\left(x\ne1;x\ne-2\right)\)

\(\Leftrightarrow\frac{3}{x^2+x-2}-\frac{1}{x-1}+\frac{7}{x+2}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{6x-8}{\left(x-1\right)\left(x+2\right)}=0\)

=> 6x-8=0

<=> x=\(\frac{8}{6}=\frac{4}{3}\left(tmđk\right)\)

b) ĐKXĐ: x khác 2; x khác 4

\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)

<=> \(\frac{2}{\left(x-2\right)\left(x-4\right)}+\frac{x-1}{x-2}=\frac{x+3}{x-4}\)

<=> 2(x - 2) + (x - 1)(x - 4)(x - 2) = (x + 3)(x - 2)(x - 2)

<=> x^3 - 7x^2 + 16x - 12 = -x^3 + x^2 + 8x - 12

<=> x^2 - 7x^2 + 16x - 12 + x^3 - x^2 + 8x - 12 = 0

<=> 2x^3 - 8x^2 + 8x = 0

<=> 2x(x - 2)(x - 2) = 0

<=> 2x = 0 hoặc x - 2 = 0

<=> x = 0 (tmđk) hoặc x = 2 (ktmđk)

=> x = 2

1. C, 2.C, 3.D ,

\(\Leftrightarrow\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\frac{\left(x-1\right)^2}{\left(x+2\right)\left(x-1\right)}\)\(\Rightarrow x\left(x+2\right)=\left(x-1\right)^2\)

                                                                                      \(\Leftrightarrow x^2+2x=x^2-2x+1\)

                                                                                       \(\Leftrightarrow4x-1=0\)\(\Leftrightarrow x=\frac{1}{4}\)

\(\frac{x}{x-1}=\frac{x-1}{x+2}\)

<=> x(x + 2) = (x - 1)²

<=> x^2 + 2x = x^2 - 2x + 1

<=> x^2 + 2x - x^2 + 2x - 1 = 0

<=> 4x - 1 = 0

<=> 4x = 1

<=> x = 1/4