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AH
Akai Haruma
Giáo viên
27 tháng 1
Lời giải:
ĐKXĐ: $x\neq -1; x\neq 0; x\neq 2$
\(Q=1+\left[\frac{x+1}{(x+1)(x^2-x+1)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2(x-2)}{x(x^2-x+1)}\)
\(=1+\left[\frac{1}{x^2-x+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right].\frac{x^2-x+1}{x-2}\)
\(=1+(\frac{2}{x^2-x+1}-\frac{2}{x+1}).\frac{x^2-x+1}{x-2}\\ =1+\frac{2}{x-2}-\frac{2(x^2-x+1)}{(x+1)(x-2)}=\frac{x}{x-2}-\frac{2x^2-2x+2}{(x+1)(x-2)}\)
\(=\frac{x(x+1)-(2x^2-2x+2)}{(x+1)(x-2)}=\frac{-x^2+3x-2}{(x+1)(x-2)}=\frac{(1-x)(x-2)}{(x+1)(x-2)}=\frac{1-x}{1+x}\)
AT
9 tháng 1 2018
Sửa đề:
\(Q=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\dfrac{x+1-x-1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\dfrac{-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x\left(x-2\right)}{x^2-x+1}\)
\(=1+\dfrac{-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\dfrac{-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-2x\right)}=\dfrac{\left(x+1\right)\left(x^2-2x\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-2x\right)}\)
\(=\dfrac{x^3-2x^2+x^2-2x-2x^2+2x-2}{\left(x+1\right)\left(x^2-2x\right)}\)
\(=\dfrac{x^3-3x^2-2}{\left(x+1\right)\left(x^2-2x\right)}\)
A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)