Đáp án:

h) \(\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{3}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{12 + 2\left( {2x - 5} \right) - 3\left( {3 - x} \right)}}{{12}} = 0\\
 \to 12 + 4x - 10 - 9 + 3x = 0\\
 \to 7x - 7 = 0\\
 \to x = 1\\
c)DK:x \ne  \pm 2\\
\dfrac{{{x^2} - 4x + 4 + 3x + 6 - {x^2} + 11}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
 \to  - x + 21 = 0\\
 \to x = 21\\
e)DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{{x^2} + 2x - x + 2 - 2}}{{x\left( {x - 2} \right)}} = 0\\
 \to {x^2} + x = 0\\
 \to x\left( {x + 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x =  - 1\left( {TM} \right)
\end{array} \right.\\
g)DK:x \ne \left\{ {1;3} \right\}\\
\dfrac{{3{x^2} - 9x - x + 3 - 2{x^2} + 2x - 5x + 5 - {x^2} + 4x - 3}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
 \to  - 9x + 5 = 0\\
 \to x = \dfrac{5}{9}\\
b)DK:x \ne \left\{ { - 1;0} \right\}\\
\dfrac{{{x^2} + 3x + \left( {x - 2} \right)\left( {x + 1} \right) - 2x\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} = 0\\
 \to {x^2} + 3x + {x^2} - x - 2 - 2{x^2} - 2x = 0\\
 \to  - 2 = 0\left( l \right)\\
 \to x \in \emptyset \\
d)DK:x \ne \left\{ { - 1;2} \right\}\\
\dfrac{{2x - 4 - x - 1 - 3x + 11}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = 0\\
 \to  - 2x + 6 = 0\\
 \to x = 3\\
f)DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{{x^2} + 2x - x + 2 - 2}}{{x\left( {x - 2} \right)}} = 0\\
 \to {x^2} + x = 0\\
 \to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x =  - 1
\end{array} \right.\\
h)DK:x \ne  \pm \dfrac{1}{2}\\
\dfrac{{4{x^2} + 2x + 2{x^2} - x - 4{x^2} + 1 - 4}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = 0\\
 \to 2{x^2} + x - 3 = 0\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{3}{2}
\end{array} \right.
\end{array}\)