*Giải phương trình chứa ẩn ở mẫu: a, (x+2/x+3)+(x-1/x)=2 b, (x^2-5x/x-5)=5 c, (x+2/x-2)-(x-2/x+2)=(3/x^2-4) d, (4x-3/x+5)=(12x+1/3x-1) GIÚP MIK VS CTLHN LUÔN !!!!

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne  - 3;x \ne 0\\
\dfrac{{x + 2}}{{x + 3}} + \dfrac{{x - 1}}{x} = 2\\
 \Leftrightarrow \dfrac{{\left( {x + 2} \right).x + \left( {x - 1} \right).\left( {x + 3} \right)}}{{x\left( {x + 3} \right)}} = 2\\
 \Leftrightarrow {x^2} + 2x + {x^2} + 2x - 3 = 2\left( {{x^2} + 3x} \right)\\
 \Leftrightarrow 2{x^2} + 4x - 3 = 2{x^2} + 6x\\
 \Leftrightarrow 2x =  - 3\\
 \Leftrightarrow x =  - \dfrac{3}{2}\left( {tm} \right)\\
Vậy\,x =  - \dfrac{3}{2}\\
b)Dk:x \ne 5\\
\dfrac{{{x^2} - 5x}}{{x - 5}} = 5\\
 \Leftrightarrow \dfrac{{x\left( {x - 5} \right)}}{{x - 5}} = 5\\
 \Leftrightarrow x = 5\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)Dkxd:x \ne 2;x \ne  - 2\\
\dfrac{{x + 2}}{{x - 2}} - \dfrac{{x - 2}}{{x + 2}} = \dfrac{3}{{{x^2} - 4}}\\
 \Leftrightarrow \dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{3}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 \Leftrightarrow {x^2} + 4x + 4 - {x^2} + 4x - 4 = 3\\
 \Leftrightarrow 8x = 3\\
 \Leftrightarrow x = \dfrac{3}{8}\left( {tm} \right)\\
Vậy\,x = \dfrac{3}{8}\\
d)Dk:x \ne  - 5;x \ne \dfrac{1}{3}\\
\dfrac{{4x - 3}}{{x + 5}} = \dfrac{{12x + 1}}{{3x - 1}}\\
 \Leftrightarrow \left( {4x - 3} \right).\left( {3x - 1} \right) = \left( {x + 5} \right).\left( {12x + 1} \right)\\
 \Leftrightarrow 12{x^2} - 4x - 9x + 3 = 12{x^2} + x + 60x + 5\\
 \Leftrightarrow  - 13x + 3 = 61x + 5\\
 \Leftrightarrow 74x =  - 2\\
 \Leftrightarrow x =  - \dfrac{1}{{37}}\left( {tm} \right)\\
Vậy\,x =  - \dfrac{1}{{37}}
\end{array}$