Toán Lớp 8: x+3/x^2-1-1/x^2+x=
kết quả (x^3+5x^2+7x-13) :(x^2+6x+13)
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Toán Lớp 8: x+3/x^2-1-1/x^2+x=
kết quả (x^3+5x^2+7x-13) :(x^2+6x+13)
Comments ( 1 )
Giải đáp:
$\begin{array}{l}
a)\\
\dfrac{{x + 3}}{{{x^2} – 1}} – \dfrac{1}{{{x^2} + x}}\\
= \dfrac{{x + 3}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} – \dfrac{1}{{x\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 3} \right).x – \left( {x – 1} \right)}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
= \dfrac{{{x^2} + 3x – x + 1}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
= \dfrac{{{x^2} + 2x + 1}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
= \dfrac{{{{\left( {x + 1} \right)}^2}}}{{x\left( {x + 1} \right)\left( {x – 1} \right)}}\\
= \dfrac{{x + 1}}{{x\left( {x – 1} \right)}}\\
b)\left( {{x^3} + 5{x^2} + 7x – 13} \right):\left( {{x^2} + 6x + 13} \right)\\
= \left( {{x^3} + 6{x^2} + 13x – {x^2} – 6x – 13} \right):\left( {{x^2} + 6x + 13} \right)\\
= \left( {{x^2} + 6x + 13} \right)\left( {x – 1} \right):\left( {{x^2} + 6x + 13} \right)\\
= x – 1
\end{array}$