a)x+9/x^2-9-3/x^2+3x b)3/2x+6-x-6/2x^2+6x c)x+1/x-3-1-x/x+3-2x(1-x)/9-x^2

Đáp án:

 c) \(\dfrac{2}{{x - 3}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{x + 9}}{{{x^2} - 9}} - \dfrac{3}{{{x^2} + 3x}}\\
 = \dfrac{{x\left( {x + 9} \right) - 3\left( {x - 3} \right)}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 9x - 3x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{{\left( {x + 3} \right)}^2}}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{x + 3}}{{x\left( {x - 3} \right)}}\\
b)\dfrac{3}{{2x + 6}} - \dfrac{{x - 6}}{{2{x^2} + 6x}}\\
 = \dfrac{{3x - x + 6}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{2x + 6}}{{x\left( {2x + 6} \right)}} = \dfrac{1}{x}\\
c)\dfrac{{x + 1}}{{x - 3}} - \dfrac{{1 - x}}{{x + 3}} - \dfrac{{2x\left( {1 - x} \right)}}{{9 - {x^2}}}\\
 = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right) - \left( {1 - x} \right)\left( {x - 3} \right) + 2x - 2{x^2}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 4x + 3 + {x^2} - 4x + 3 + 2x - 2{x^2}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{2x + 6}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{2}{{x - 3}}
\end{array}\)