Toán Lớp 8: A = ( x/(x-1)+ 3/(x+1) + (3-5x)/(x^2-1) ) : (-x)/3 (với x ≠ 0; x ≠ -1; x ≠ 1)
a, rút gọn A
b, tính a khi x = -2
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Toán Lớp 8: A = ( x/(x-1)+ 3/(x+1) + (3-5x)/(x^2-1) ) : (-x)/3 (với x ≠ 0; x ≠ -1; x ≠ 1)
a, rút gọn A
b, tính a khi x = -2
Comments ( 1 )
a) – \dfrac{3}{{x + 1}}\\
b)3
\end{array}\)
a)A = \left( {\dfrac{x}{{x – 1}} + \dfrac{3}{{x + 1}} + \dfrac{{3 – 5x}}{{{x^2} – 1}}} \right):\dfrac{{ – x}}{3}\\
= \dfrac{{x\left( {x + 1} \right) + 3\left( {x – 1} \right) + 3 – 5x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{3}{{ – x}}\\
= \dfrac{{{x^2} + x + 3x – 3 + 3 – 5x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{3}{{ – x}}\\
= \dfrac{{{x^2} – x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{3}{{ – x}}\\
= \dfrac{{x\left( {x – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{3}{{ – x}}\\
= – \dfrac{3}{{x + 1}}\\
b)Thay:x = – 2\\
\to A = – \dfrac{3}{{ – 2 + 1}} = 3
\end{array}\)