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\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)
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A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
27 tháng 2 2020
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow A=\left(\frac{-1}{x-1}+\frac{2}{x+1}+\frac{5-x}{x^2-1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\left[\frac{-x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-2}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Leftrightarrow A=\frac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}\)
\(\Leftrightarrow A=\frac{2\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=1\)
vậy \(A=1\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)
A
13 tháng 8 2020
\(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1}{\left(1-x\right)\left(x+1\right)}+\frac{2\left(1-x\right)}{\left(x+1\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}.\)
\(A=\frac{2}{x^2-1}\cdot\frac{^2-1}{1-2x}=\frac{2}{1-2x}\)ĐK: x khác 1/2
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
19 tháng 3 2020
=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]
={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]
=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]
=-x.x(x-2)(x+1)/2(2-x)x3
=(x+1)/2x
\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{x^2-1}.\frac{x^2-1}{5}=\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)
\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\left(x\ne\pm1\right)\)
\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{\left(x^2+5x+8\right)\cdot\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)5}=\frac{x^2+5x+8}{5}\)