a)3/x^2+x-2-1/x-=-7/x+2 b)5/-x^2+5x-6+x+3/2-x=0 c)1/x-1-3x^2/x^3-1=2x/x^2+x+1

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne 1;x \ne  - 2\\
\dfrac{3}{{{x^2} + x - 2}} - \dfrac{1}{{x - 1}} =  - \dfrac{7}{{x + 2}}\\
 \Leftrightarrow \dfrac{{3 - \left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{{ - 7\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\\
 \Leftrightarrow 3 - x - 2 =  - 7x + 7\\
 \Leftrightarrow 6x = 6\\
 \Leftrightarrow x = 1\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
b)Dkxd:x \ne 2;x \ne 3\\
\dfrac{5}{{ - {x^2} + 5x - 6}} + \dfrac{{x + 3}}{{2 - x}} = 0\\
 \Leftrightarrow  - \dfrac{5}{{{x^2} - 5x + 6}} - \dfrac{{x + 3}}{{x - 2}} = 0\\
 \Leftrightarrow \dfrac{{5 + \left( {x + 3} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = 0\\
 \Leftrightarrow 5 + {x^2} - 9 = 0\\
 \Leftrightarrow {x^2} = 4\\
 \Leftrightarrow x =  - 2\left( {do:x \ne 2} \right)\\
Vậy\,x =  - 2\\
c)Dkxd:x \ne 1\\
\dfrac{1}{{x - 1}} - \dfrac{{3{x^2}}}{{{x^3} - 1}} = \dfrac{{2x}}{{{x^2} + x + 1}}\\
 \Leftrightarrow \dfrac{{{x^2} + x + 1 - 3{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{2x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 \Leftrightarrow  - 2{x^2} + x + 1 = 2{x^2} - 2x\\
 \Leftrightarrow 4{x^2} - 3x - 1 = 0\\
 \Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right)\\
 \Leftrightarrow x =  - \dfrac{1}{4}\left( {do:x \ne 1} \right)\\
Vậy\,x =  - \dfrac{1}{4}
\end{array}$