Toán Lớp 8: Giúp mình thực hiện phép tính này với ạ.
7 $\frac{7}{8x^2-18}$ + $\frac{1}{2x^2+3x}$ – $\frac{1}{4x-6}$
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Toán Lớp 8: Giúp mình thực hiện phép tính này với ạ.
7 $\frac{7}{8x^2-18}$ + $\frac{1}{2x^2+3x}$ – $\frac{1}{4x-6}$
Comments ( 2 )
\dfrac{7}{{8{x^2} – 18}} + \dfrac{1}{{2{x^2} + 3x}} – \dfrac{1}{{4x – 6}}\\
= \dfrac{7}{{2\left( {4{x^2} – 9} \right)}} + \dfrac{1}{{x\left( {2x + 3} \right)}} – \dfrac{1}{{2\left( {2x – 3} \right)}}\\
= \dfrac{7}{{2\left( {2x – 3} \right)\left( {2x + 3} \right)}} + \dfrac{1}{{x\left( {2x + 3} \right)}} – \dfrac{1}{{2\left( {2x – 3} \right)}}\\
MC:2x\left( {2x – 3} \right)\left( {2x + 3} \right)\\
= \dfrac{{7x}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}} + \dfrac{{2\left( {2x – 3} \right)}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}} – \dfrac{{x\left( {2x + 3} \right)}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{{7x + 2\left( {2x – 3} \right) – x\left( {2x + 3} \right)}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{{7x + 4x – 6 – 2{x^2} – 3x}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{{ – 2{x^2} + 8x – 6}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{{ – {x^2} + 4x – 3}}{{2x\left( {2x – 3} \right)\left( {2x + 3} \right)}}
\end{array}\]