Đáp án:
e) \(\dfrac{1}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
8)a)\dfrac{{5.6.y + 7.3.x + 2xy}}{{36{x^2}{y^2}}} = \dfrac{{30y + 21x + 2xy}}{{36{x^2}{y^2}}}\\
b)\dfrac{{1 - 2x + 3 + 2y + 2x - 4}}{{6{x^3}y}}\\
= \dfrac{{2y}}{{6{x^3}y}} = \dfrac{1}{{3{x^3}}}\\
c)\dfrac{{{x^2} - 2 + 2 - x}}{{x{{\left( {x - 1} \right)}^2}}} = \dfrac{{x\left( {x - 1} \right)}}{{x{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{1}{{x - 1}}\\
d)\dfrac{3}{{2x}} + \dfrac{{3x - 3}}{{2x - 1}} + \dfrac{{2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{3\left( {2x - 1} \right) + 2x\left( {3x - 3} \right) + 2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{6x - 3 + 6{x^2} - 6x + 2{x^2} + 1}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{8{x^2} - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{{2\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{2x\left( {2x - 1} \right)}}\\
= \dfrac{{2x - 1}}{x}\\
e)\dfrac{{4\left( {x - 2} \right) + 2\left( {x + 2} \right) - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x - 8 + 2x + 4 - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x - 2}}
\end{array}\)
