Đáp án:
a) \(\dfrac{{x + 2}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{3x + 2 - 4\left( {3x - 2} \right) + 10x - 8}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{{13x - 6 - 12x + 8}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{{x + 2}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
b)\dfrac{3}{{2x\left( {x + 1} \right)}} + \dfrac{{2x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{2}{x}\\
= \dfrac{{3\left( {x - 1} \right) + 2x\left( {2x - 1} \right) - 4\left( {{x^2} - 1} \right)}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x - 3 + 4{x^2} - 2x - 4{x^2} + 4}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x + 1}}{{2x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{2x\left( {x - 1} \right)}}\\
c)\dfrac{{3x}}{{5\left( {x + y} \right)}} - \dfrac{x}{{10\left( {x - y} \right)}}\\
= \dfrac{{6x\left( {x - y} \right) - x\left( {x + y} \right)}}{{10\left( {x - y} \right)\left( {x + y} \right)}}\\
= \dfrac{{6{x^2} - 6xy - {x^2} - xy}}{{10\left( {x - y} \right)\left( {x + y} \right)}}\\
= \dfrac{{5{x^2} - 7xy}}{{10\left( {x - y} \right)\left( {x + y} \right)}}
\end{array}\)
