Giải giúp mik

1) 11x+13/3x-3+15x+17/4x-4

2) 3x+5/x^2-5x+25-x/25-5x

3) 5/4*(x+2)+8-x/4x^2+8x

4) xy/x^2-y^2-x^2/y^2-x^2

Đáp án:

$\begin{array}{l}
1)Dkxd:x \ne 1\\
\dfrac{{11x + 13}}{{3x - 3}} + \dfrac{{15x + 17}}{{4x - 4}}\\
 = \dfrac{{4.\left( {11x + 13} \right) + 3.\left( {15x + 17} \right)}}{{12.\left( {x - 1} \right)}}\\
 = \dfrac{{89x + 103}}{{12\left( {x - 1} \right)}}\\
2)Dkxd:x \ne 0;x \ne 5\\
\dfrac{{3x + 5}}{{{x^2} - 5x}} + \dfrac{{25 - x}}{{25 - 5x}}\\
 = \dfrac{{5.\left( {3x + 5} \right) - x.\left( {25 - x} \right)}}{{5x\left( {x - 5} \right)}}\\
 = \dfrac{{15x + 25 - 25x + {x^2}}}{{5x\left( {x - 5} \right)}}\\
 = \dfrac{{{x^2} - 10x + 25}}{{5x\left( {x - 5} \right)}}\\
 = \dfrac{{x - 5}}{{5x}}\\
3)Dkxd:x \ne 0;x \ne  - 2\\
\dfrac{5}{{4\left( {x + 2} \right)}} + \dfrac{{8 - x}}{{4{x^2} + 8x}}\\
 = \dfrac{{5x + 8 - x}}{{4x\left( {x + 2} \right)}}\\
 = \dfrac{{4x + 8}}{{x\left( {4x + 8} \right)}}\\
 = \dfrac{1}{x}\\
4)Dk:x \ne y;x \ne  - y\\
\dfrac{{xy}}{{{x^2} - {y^2}}} - \dfrac{{{x^2}}}{{{y^2} - {x^2}}}\\
 = \dfrac{{xy + {x^2}}}{{{x^2} - {y^2}}}\\
 = \dfrac{{x\left( {y + x} \right)}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
 = \dfrac{x}{{x - y}}
\end{array}$