Đáp án:
a) \(\dfrac{2}{{x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{4}{{x + 2}} + \dfrac{3}{{x - 2}} + \dfrac{{5x + 2}}{{4 - {x^2}}}\\
= \dfrac{{4\left( {x - 2} \right) + 3\left( {x + 2} \right) - 5x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x - 8 + 3x + 6 - 5x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{2x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{2}{{x + 2}}\\
b)\dfrac{{1 + x}}{{x - 3}} + \dfrac{{1 - 2x}}{{3 + x}} + \dfrac{{x\left( {1 - x} \right)}}{{9 - {x^2}}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + \left( {1 - 2x} \right)\left( {x - 3} \right) - x + {x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 1 - 2{x^2} + 7x - 3 - x + {x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{6x - 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
c)\dfrac{{3x + 2}}{{{x^2} - 2x + 1}} + \dfrac{6}{{{x^2} - 1}} + \dfrac{{3x - 2}}{{{x^2} + 2x + 1}}\\
= \dfrac{{\left( {3x + 2} \right)\left( {{x^2} + 2x + 1} \right) + 6\left( {{x^2} - 1} \right) + \left( {3x - 2} \right)\left( {{x^2} - 2x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{3{x^3} + 6{x^2} + 3x + 2{x^2} + 4x + 2 + 6{x^2} - 6 + 3{x^3} - 6{x^2} + 3x - 2{x^2} + 4x - 2}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{6{x^3} + 6{x^2} + 14x - 6}}{{{{\left( {x + 1} \right)}^2}.{{\left( {x - 1} \right)}^2}}}
\end{array}\)
