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12 tháng 4 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{1}{2}\\x\ne-\frac{1}{2}\\x\ne0\end{matrix}\right.\)
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\left[1:\left(1-\frac{1}{x}+\frac{1}{4x^2}\right)\right]\)
\(=\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\left[1:\frac{4x^2-4x+1}{4x^2}\right]\)
\(=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x^2}{\left(2x-1\right)^2}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{\left(2x-1\right)^2}{4x^2}=\frac{2\left(2x-1\right)}{\left(2x+1\right).x}=\frac{4x-2}{2x^2+x}\left(ĐPCM\right)\)
2 tháng 2 2022
\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)
\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)
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=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)
NT
24 tháng 8 2016
\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)
24 tháng 8 2016
\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-4^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)
15 tháng 12 2014
Cái này nhanh hơn nàk bạn:
a/ Đặt a=2x-1,b=x+1, biểu thứ trở thành:
a2 +2ab+b2
=(a+b)2
=>(2x-1+x+1)2=(3x)2=9x2
\(\left(2x+1-\frac{1}{1-2x}\right):\left(2-\frac{4x}{2x-1}\right)\)
\(=\frac{\left(2x+1\right)\left(2x-1\right)+1}{2x-1}:\frac{4x-2-4x}{2x-1}\)
\(=\frac{\left(2x+1\right)\left(2x-1\right)+1}{-2}\)
\(=\frac{\left(4x^2-1\right)+1}{-2}=\frac{4x^2}{-2}=-2x^2\)
\(\left(2x+1-\frac{1}{1-2x}\right):\left(2-\frac{4x}{2x-1}\right)\)
\(\Leftrightarrow\left(\frac{\left(2x-1\right)\left(2x+1\right)+1}{2x-1}\right):\left(\frac{2\left(2x-1\right)-4x}{2x-1}\right)\)
\(\Leftrightarrow\left(\frac{4x^2-1+1}{2x-1}\right):\left(\frac{4x-2-4x}{2x-1}\right)\)
\(\Leftrightarrow\frac{4x^2}{2x-1}.\frac{2x-1}{-2}\Leftrightarrow\frac{4x^2}{-2}\Leftrightarrow-2x^2\)