Rút gọn : \(\left(\frac{1+x}{x}+\frac{1}{4x^2}\right)\left(\frac{1-2x}{1+2x}-\frac{1}{1-4x^2}\times\frac{1-4x+4x^2}{1+2x}\right)-\frac{1}{2x}\)
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6 tháng 7 2016
mk ko biết cứ bấm đại thui, bn có thể giúp mk ko ???
31 tháng 1 2022
\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}=\dfrac{x^2+1}{2}\)
DT
1
29 tháng 1 2022
\(=\dfrac{3x+6x^2+2x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\cdot\dfrac{\left(1-2x\right)^2}{x\left(2x+5\right)}\)
\(=\dfrac{1-2x}{1+2x}\)
18 tháng 6 2016
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right):\frac{4}{4x^2-4}\)
\(=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+2\right)}+\frac{6}{2.\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\frac{4}{4\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)=\frac{4}{2}=2\)
12 tháng 4 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{1}{2}\\x\ne-\frac{1}{2}\\x\ne0\end{matrix}\right.\)
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\left[1:\left(1-\frac{1}{x}+\frac{1}{4x^2}\right)\right]\)
\(=\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\left[1:\frac{4x^2-4x+1}{4x^2}\right]\)
\(=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x^2}{\left(2x-1\right)^2}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{\left(2x-1\right)^2}{4x^2}=\frac{2\left(2x-1\right)}{\left(2x+1\right).x}=\frac{4x-2}{2x^2+x}\left(ĐPCM\right)\)
22 tháng 2 2020
ĐKXĐ : \(x\ne\pm\frac{1}{2}\)
\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)
\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)
\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)
\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)
\(E=\frac{8x^3+1}{1+4x^2}\)
Study well
LT
22 tháng 2 2020
E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)
E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)
E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)
E=\(\frac{4x^3+1}{1+4x^2}\)
22 tháng 2 2020
E=\(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{1-4x^2}\)
E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^2+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{1+4x+4x^2-1+4x-4x^2}\)
E=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)
\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)
\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)