Toán Lớp 8: 1/ x^2 + x + 2x – 5 / 2x^2 – 2
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Toán Lớp 8: 1/ x^2 + x + 2x – 5 / 2x^2 – 2
Comments ( 2 )
\dfrac{1}{{{x^2} + x}} + \dfrac{{2x – 5}}{{2{x^2} – 2}}\\
= \dfrac{1}{{x\left( {x + 1} \right)}} + \dfrac{{2x – 5}}{{2\left( {{x^2} – 1} \right)}}\\
= \dfrac{1}{{x\left( {x + 1} \right)}} + \dfrac{{2x – 5}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
MC:2x\left( {x – 1} \right)\left( {x + 1} \right)\\
= \dfrac{{2\left( {x – 1} \right)}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}} + \dfrac{{x\left( {2x – 5} \right)}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2x – 2}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}} + \dfrac{{2{x^2} – 5x}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2x – 2 + 2{x^2} – 5x}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2{x^2} – 3x – 2}}{{2x\left( {x – 1} \right)\left( {x + 1} \right)}}
\end{array}\]