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27 tháng 6 2016
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
31 tháng 3 2020
\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
1 tháng 4 2020
ĐKXD: x\(\ne\)-1,-2,-3
Ta có
\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
Chúc bạn học tốt
\(\frac{\left(x-1\right)^2}{4}+x^2-1+\left(x+1\right)^2=\frac{x^2}{4}-\frac{x}{2}+\frac{1}{4}+x^2-1+x^2+2x+1\)
\(=\frac{9}{4}x^2+\frac{3}{2}x+\frac{1}{4}=\frac{\left(3x+1\right)^2}{4}\)