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 

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=\(\frac{4x^3+1}{1+4x^2}\)

Góp ý : 

Cô giáo ơi

Cô tính nhầm rồi 2x + 2x = 4x cô ạ chứ không phải 8x đâu cô

mong cô xem lại ạ 

\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)

\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)

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=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{1}{2}\\x\ne-\frac{1}{2}\\x\ne0\end{matrix}\right.\)

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\left[1:\left(1-\frac{1}{x}+\frac{1}{4x^2}\right)\right]\)

\(=\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\left[1:\frac{4x^2-4x+1}{4x^2}\right]\)

\(=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}:\frac{4x^2}{\left(2x-1\right)^2}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{\left(2x-1\right)^2}{4x^2}=\frac{2\left(2x-1\right)}{\left(2x+1\right).x}=\frac{4x-2}{2x^2+x}\left(ĐPCM\right)\)

\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

\(=\frac{1}{2x}-1+1+\frac{1}{2x-1}+\frac{1}{2x\left(1-2x\right)}=\frac{1-2x}{2x\left(1-2x\right)}-\frac{2x}{2x\left(1-2x\right)}+\frac{1}{2x\left(1-2x\right)}\)

\(=\frac{1-2x-2x+1}{2x\left(1-2x\right)}=\frac{2}{2x\left(1-2x\right)}=\frac{1}{x\left(1-2x\right)}\)

Ta có: \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

= \(\frac{1-2x}{2x}+\frac{2x}{2x-1}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{2x.2x}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-\left(4x^2-4x+1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{2x\left(2x-1\right)}-\frac{1}{2x\left(2x-1\right)}\)

= \(\frac{-4x^2+4x-1+4x^2-1}{2x\left(2x-1\right)}\)

= \(\frac{4x-2}{2x\left(2x-1\right)}\)

= \(\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)

\(=\dfrac{3x+6x^2+2x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\cdot\dfrac{\left(1-2x\right)^2}{x\left(2x+5\right)}\)

\(=\dfrac{1-2x}{1+2x}\)

\(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}=\frac{8}{4x^2-1}\)

\(\Leftrightarrow\frac{\left(2x+1\right)^2}{4x^2-1}-\frac{\left(2x-1\right)^2}{4x^2-1}=\frac{8}{4x^2-1}\)

\(\Leftrightarrow\frac{4x^2+4x+1-4x^2+4x-1-8}{4x^2-1}=0\)

\(\Leftrightarrow\frac{8x-8}{4x^2-1}=0\)

\(\Rightarrow8x-8=0\)

\(\Rightarrow x=1\)

tick mình nha!

\(\Leftrightarrow\frac{\left(2x+1\right)^2}{4x^2-1}-\frac{\left(2x-1\right)^2}{4x^2-1}=\frac{9}{4x^2-1}\)

\(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2+4x+1=9\)

\(\Leftrightarrow8x=7\)

Vậy x=7/8