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3 tháng 6 2015

Ta có: \(\frac{a+1}{2a-2}-\frac{1}{2a^2-2}=\frac{\left(a+1\right)^2-1}{2\left(a^2-1\right)}=\frac{a^2+2a+1-1}{2\left(a^2-1\right)}=\frac{a\left(a+2\right)}{2\left(a^2-1\right)}\)

Vậy D=\(\frac{a\left(a+2\right)}{2\left(a^2-1\right)}.\frac{2\left(a+1\right)}{a+2}=\frac{a}{a-1}\)

\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)

16 tháng 3 2020

\(\text{GIẢI :}\)

ĐKXĐ : \(a\ne\pm1\).

\(M=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a^3-a}\right):\frac{a^2-2a+1}{a+a^3}\)

\(=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)

\(=\frac{1}{a^2-2a+1}-\left(\frac{a^2}{a\left(a^2-1\right)}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)

\(=\frac{1}{a^2-2a+1}-\frac{a^2-1}{a\left(a^2-1\right)}:\frac{\left(a-1\right)^2}{a\left(1+a^2\right)}\)

\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{a\left(a^2-1\right)}\cdot\frac{a\left(a^2+1\right)}{1+a^2}\)

\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{1+a^2}=\frac{-a^2}{\left(a-1\right)^2}\).

30 tháng 9 2016

\(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\left(ĐK:2a\ne\pm b\right)\)

\(=\left(\frac{1}{2a-b}-\frac{3b}{\left(2b-b\right)\left(2a+b\right)}-\frac{2}{2a+b}\right):\frac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)

\(=\frac{2a+b-3b-2\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\cdot\frac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)

\(=\frac{2a+b-3b-4a+2b}{8a^2}=\frac{-2a}{8a^2}=-\frac{1}{4a}\)