Toán Lớp 8: Tính và rút gọn :
a) x+1/x²-2x-8 . 4-x/x²+x
b) x+2/4x+24 . x²-36/x²+x-2
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Toán Lớp 8: Tính và rút gọn :
a) x+1/x²-2x-8 . 4-x/x²+x
b) x+2/4x+24 . x²-36/x²+x-2
Comments ( 1 )
Giải đáp:
$\begin{array}{l}
a)\dfrac{{x + 1}}{{{x^2} – 2x – 8}}.\dfrac{{4 – x}}{{{x^2} + x}}\\
= \dfrac{{x + 1}}{{\left( {x – 4} \right)\left( {x + 2} \right)}}.\dfrac{{ – \left( {x – 4} \right)}}{{x\left( {x + 1} \right)}}\\
= \dfrac{1}{{x + 2}}.\dfrac{{ – 1}}{x}\\
= \dfrac{{ – 1}}{{x\left( {x + 2} \right)}}\\
b)\dfrac{{x + 2}}{{4x + 24}}.\dfrac{{{x^2} – 36}}{{{x^2} + x – 2}}\\
= \dfrac{{x + 2}}{{4.\left( {x + 6} \right)}}.\dfrac{{\left( {x + 6} \right)\left( {x – 6} \right)}}{{\left( {x – 1} \right)\left( {x + 2} \right)}}\\
= \dfrac{1}{4}.\dfrac{{x – 6}}{{x – 1}}\\
= \dfrac{{x – 6}}{{4x – 4}}
\end{array}$