Rút gọn A=\(\left(\frac{a^3-3}{a^2-2a-3}-\frac{2a-6}{a+1}+\frac{a+3}{5-a}\right):\frac{a^2+8}{a^2-1}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{GIẢI :}\)
ĐKXĐ : \(a\ne\pm1\).
\(M=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a^3-a}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a^2}{a\left(a^2-1\right)}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\frac{a^2-1}{a\left(a^2-1\right)}:\frac{\left(a-1\right)^2}{a\left(1+a^2\right)}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{a\left(a^2-1\right)}\cdot\frac{a\left(a^2+1\right)}{1+a^2}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{1+a^2}=\frac{-a^2}{\left(a-1\right)^2}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)
\(=\left(\frac{a^2+1}{a^2+1}+\frac{a}{a^2+1}\right):\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{\left(a^2+1\right)\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\frac{a-1}{a^2+1}=\frac{a^2+a+1}{a-1}\)
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Trả lời
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\) \(\left(a\ge0.a\ne1\right)\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{1}{\left(a+1\right)^2}-\frac{1}{\left(a-1\right).\left(a+1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{a-1-a-1}{\left(a+1\right)^2.\left(a-1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.0\)
\(B=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}-\frac{a^2+1}{\left(a^2-1\right)\left(a^2+1\right)}\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1-a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}\right)\)
\(=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)