$\frac{12x+5}{3}$=$\frac{2x-7}{4}$ $\frac{3(x-11)}{4}$=$\frac{3(x+1)}{5}$-$\frac{2(2x-5)}{10}$ $x^{2}$ -7x+6=0 $\frac{x+5}{x-1}$=$\frac{x+1}{x-3}$-$\frac{8}{ x^{2} -4x+3}$ $\frac{x+1}{x-2}$-$\frac{5}{x+2}$=$\frac{12}{ x^{2} -4}$+1

Đáp án:

$\begin{array}{l}
a)\dfrac{{12x + 5}}{3} = \dfrac{{2x - 7}}{4}\\
 \Leftrightarrow 4.\left( {12x + 5} \right) = 3.\left( {2x - 7} \right)\\
 \Leftrightarrow 48x + 20 = 6x - 21\\
 \Leftrightarrow 42x =  - 41\\
 \Leftrightarrow x =  - \dfrac{{41}}{{42}}\\
Vậy\,x =  - \dfrac{{41}}{{42}}\\
b)\dfrac{{3\left( {x - 11} \right)}}{4} = \dfrac{{3\left( {x + 1} \right)}}{5} - \dfrac{{2\left( {2x - 5} \right)}}{{10}}\\
 \Leftrightarrow \dfrac{{3\left( {x - 11} \right)}}{4} = \dfrac{{3x + 3 - 2x + 5}}{5}\\
 \Leftrightarrow \dfrac{{3\left( {x - 11} \right)}}{4} = \dfrac{{x + 8}}{5}\\
 \Leftrightarrow 15.\left( {x - 11} \right) = 4.\left( {x + 8} \right)\\
 \Leftrightarrow 15x - 161 = 4x + 32\\
 \Leftrightarrow 11x = 193\\
 \Leftrightarrow x = \dfrac{{193}}{{11}}\\
Vậy\,x = \dfrac{{193}}{{11}}\\
c){x^2} - 7x + 6 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 6} \right) = 0\\
 \Leftrightarrow x = 1;x = 6\\
Vậy\,x = 1;x = 6\\
d)Dk:x \ne 1;x \ne 3\\
\dfrac{{x + 5}}{{x - 1}} = \dfrac{{x + 1}}{{x - 3}} - \dfrac{8}{{{x^2} - 4x + 3}}\\
 \Leftrightarrow \dfrac{{\left( {x + 5} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) - 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
 \Leftrightarrow {x^2} + 2x - 15 = {x^2} - 1 - 8\\
 \Leftrightarrow 2x = 6\\
 \Leftrightarrow x = 3\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
e)Dk:x \ne 2;x \ne  - 2\\
\dfrac{{x + 1}}{{x - 2}} - \dfrac{5}{{x + 2}} = \dfrac{{12}}{{{x^2} - 4}} + 1\\
 \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) - 5\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{12 + {x^2} - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 \Leftrightarrow {x^2} + 3x + 2 - 5x + 10 = 8 + {x^2}\\
 \Leftrightarrow  - 2x + 12 = 8\\
 \Leftrightarrow 2x = 4\\
 \Leftrightarrow x = 2\left( {ktm} \right)\\
Vậy\,x \in \emptyset 
\end{array}$