Toán Lớp 8: a)(2-x).(2+x)/x^3+8+x-2/x^2-2x+4+2x/x+2 b)7x+1/x-3-2.(x-14)/3-x
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Toán Lớp 8: a)(2-x).(2+x)/x^3+8+x-2/x^2-2x+4+2x/x+2 b)7x+1/x-3-2.(x-14)/3-x
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a)\\
\dfrac{{\left( {2 – x} \right)\left( {2 + x} \right)}}{{{x^3} + 8}} + \dfrac{{x – 2}}{{{x^2} – 2x + 4}} + \dfrac{{2x}}{{x + 2}}\\
= \dfrac{{4 – {x^2}}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}} + \dfrac{{x – 2}}{{{x^2} – 2x + 4}} + \dfrac{{2x}}{{x + 2}}\\
= \dfrac{{4 – {x^2} + \left( {x – 2} \right)\left( {x + 2} \right) + 2x.\left( {{x^2} – 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}}\\
= \dfrac{{4 – {x^2} + {x^2} – 4 + 2{x^3} – 4{x^2} + 8x}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}}\\
= \dfrac{{2{x^3} – 4{x^2} + 8x}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}}\\
= \dfrac{{2x\left( {{x^2} – 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}}\\
= \dfrac{{2x}}{{x + 2}}\\
b)\\
\dfrac{{7x + 1}}{{x – 3}} – \dfrac{{2\left( {x – 14} \right)}}{{3 – x}}\\
= \dfrac{{7x + 1 + 2\left( {x – 14} \right)}}{{x – 3}}\\
= \dfrac{{7x + 1 + 2x – 28}}{{x – 3}}\\
= \dfrac{{9x – 27}}{{x – 3}}\\
= \dfrac{{9\left( {x – 3} \right)}}{{x – 3}}\\
= 9
\end{array}$