Toán Lớp 8: 4x-4/2x^2+6x;x-3/5x^2+10 quy đồng mẫu
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Toán Lớp 8: 4x-4/2x^2+6x;x-3/5x^2+10 quy đồng mẫu
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\dfrac{{5{x^3} – 10{x^2} + 20x – 20}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}\\
\dfrac{{{x^3} – 9x}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}
\end{array}\)
\dfrac{{4x – 4}}{{2{x^2} + 6x}} = \dfrac{{4\left( {x – 1} \right)}}{{2x\left( {x + 3} \right)}} = \dfrac{{2x – 2}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{\left( {2x – 2} \right)\left( {5{x^2} + 10} \right)}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}\\
= \dfrac{{5{x^3} – 10{x^2} + 20x – 20}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}\\
\dfrac{{x – 3}}{{5{x^2} + 10}} = \dfrac{{x – 3}}{{5\left( {{x^2} + 2} \right)}}\\
= \dfrac{{\left( {x – 3} \right)\left( {{x^2} + 3x} \right)}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}\\
= \dfrac{{{x^3} – 3{x^2} + 3{x^2} – 9x}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}\\
= \dfrac{{{x^3} – 9x}}{{5x\left( {x + 3} \right)\left( {{x^2} + 2} \right)}}
\end{array}\)