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\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
\(\text{GIẢI :}\)
ĐKXĐ : \(a\ne\pm1\).
\(M=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a^3-a}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a^2}{a\left(a^2-1\right)}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\frac{a^2-1}{a\left(a^2-1\right)}:\frac{\left(a-1\right)^2}{a\left(1+a^2\right)}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{a\left(a^2-1\right)}\cdot\frac{a\left(a^2+1\right)}{1+a^2}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{1+a^2}=\frac{-a^2}{\left(a-1\right)^2}\).
\(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\left(ĐK:2a\ne\pm b\right)\)
\(=\left(\frac{1}{2a-b}-\frac{3b}{\left(2b-b\right)\left(2a+b\right)}-\frac{2}{2a+b}\right):\frac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\frac{2a+b-3b-2\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\cdot\frac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)
\(=\frac{2a+b-3b-4a+2b}{8a^2}=\frac{-2a}{8a^2}=-\frac{1}{4a}\)
Sửa lại đề bài: 1 / 2a- b
( MÁY MK KO ĐÁNH ĐC PHÂN SỐ MONG BN THÔNG CẢM)
mới lm đc nhé bn!
a) ĐKXĐ: bn tự lm nhé !
bn biến đổi: 2a3-b+2a-a2b = (2a-b) + ( 2a3-a2b) = (2a-b) + a2(2a-b) = (2a-b)(a2+1)
rồi bn nhân 1 / 2a+b với a2+1 rồi trừ 2 phân thức với nhau sẽ ra 0 => A=0
\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)
\(=\left(\frac{a^2+1}{a^2+1}+\frac{a}{a^2+1}\right):\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{\left(a^2+1\right)\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\frac{a-1}{a^2+1}=\frac{a^2+a+1}{a-1}\)
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=\(\frac{a\left(a+2\right)}{2\left(a^2-1\right)}.\frac{2\left(a+1\right)}{a+2}=\frac{a}{a-1}\)