a: Ta có: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right):\left(\dfrac{x-1}{\sqrt{x}}\right)\)

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

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

b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-2x^2+4x-2=0\)

\(\Leftrightarrow2x^2+4x-2=0\)

\(\Leftrightarrow x^2+2x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)

c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\)

Thay x=2 vào E, ta được:

\(E=\dfrac{4\cdot2^2}{1}=16\)

a, \(E=\left(\frac{x^2+4}{x^2-4}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{x^2+4}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\left(\frac{x^2+4-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{6}{x+2}\right)\)

\(=\frac{x^2+4-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{x^2-x-2}{6\left(x-2\right)}=\frac{x+1}{6}\)

b, Ta có : \(\left|2x-3\right|=1\Leftrightarrow\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(ktmđk\right)\\x=1\end{cases}}}\)

Thay x = 1 vào biểu thức E ta được : \(\frac{1+1}{6}=\frac{2}{6}=\frac{1}{3}\)

Vậy với x = 1 thì E = 1/3 

c, Ta có : \(E< 0\)hay \(\frac{x+1}{6}< 0\Rightarrow x+1>0\)( do 6 > 0 )

\(\Leftrightarrow x>-1\)

Với với x > -1 thì E < 0 

d, Ta có E = 3 - x hay \(\frac{x+1}{6}=3-x\Rightarrow x+1=18-6x\Leftrightarrow7x=17\Leftrightarrow x=\frac{17}{7}\)

Cô hướng dẫn nhé :) 

a. ĐK: \(x>0;x\ne1\) 

Ta có \(E=\frac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4\sqrt{x}\left(x-1\right)}{x-1}:\frac{x-1}{\sqrt{x}}\)

\(\Leftrightarrow E=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}=\frac{4x^2}{\left(x-1\right)^2}\)

b. Để \(E=2\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Leftrightarrow2x^2+4x-2=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}-1\\x=-\sqrt{2}-1\left(L\right)\end{cases}}\)

c. \(x=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=2\)

Vậy E = 16.

a)Rút gọn E ta đc:

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

b)Với E=2\(\Leftrightarrow\)\(\frac{4x^2+\sqrt{x}\left(2x+2\right)-4x}{x^2-2x+1}=2\)

\(\Leftrightarrow\frac{4x^2}{x^2-2x+1}+\frac{2\sqrt{x^3}}{x^2-2x+1}-\frac{4x}{x^2-2x+1}+\frac{2\sqrt{x}}{x^2-2x+1}-2=0\)

\(\Leftrightarrow\frac{2\left(x^2\sqrt{x^3}+\sqrt{x}-1\right)}{x^2-2x+1}=0\)

\(\Leftrightarrow x^2+\sqrt{x^3}+\sqrt{x}-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{-\sqrt{x^3}-\sqrt{x}+1}=0\left(tm\right)\\\sqrt{-\sqrt{x^3}-\sqrt{x}+1}+x=0\left(loai\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\sqrt{5}-3=0\left(loai\right)\\2x+\sqrt{5}-3=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow x=-\frac{\sqrt{5}-3}{2}\left(tm\right)\)

E rút gọn E ra \(\frac{4x^2}{x^2-2x+1}\)

a: Sửa đề: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}+4\sqrt{x}\right):\dfrac{x-1}{\sqrt{x}}\)

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

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

b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow4x^2-2x^2+4x-2=0\)

\(\Leftrightarrow2x^2+4x-2=0\)

\(\Leftrightarrow x^2+2x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{2}\\x+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(nhận\right)\\x=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\)

c: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

Thay x=2 vào E, ta được: 

\(E=\dfrac{4\cdot2^2}{\left(2-1\right)^2}=16\)

Bài 1 : 

a, \(A=\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)

b, Ta có : \(\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

TH1 : Thay x = 2 vào biểu thức trên ta được : 

\(\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)

TH2 : Thay x = -2 vào biểu thức trên ta được : 

\(\frac{2}{-2+2}=\frac{2}{0}\)vô lí 

c, ta có A = 2 hay \(\frac{2}{x+2}=2\)ĐK : \(x\ne-2\)

\(\Rightarrow2x+4=2\Leftrightarrow2x=-2\Leftrightarrow x=-1\)

Vậy với x = -1 thì A = 2 

d, Ta có A < 0 hay \(\frac{2}{x+2}< 0\)

\(\Rightarrow x+2< 0\)do 2 > 0 

\(\Leftrightarrow x< -2\)

Vậy với A < 0 thì x < -2 

e, Để A nhận giá trị nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

2.

ĐKXĐ : \(x\ne\pm2\)

a. \(B=\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

b. | x - 1 | = 2 <=>\(\hept{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)<=>\(\hept{\begin{cases}x=3\\x=-1\end{cases}}\)

Với x = 3 thì \(B=\frac{3-2}{3+2}=\frac{1}{5}\)

Với x = - 1 thì \(B=\frac{-1-2}{-1+2}=-3\)

Vậy với | x - 1 | = 2 thì B đạt được 2 giá trị là B = 1/5 hoặc B = - 3

c. \(B=\frac{x-2}{x+2}=-1\)<=>\(-\left(x-2\right)=x+2\)

<=> \(-x+2=x+2\)<=>\(-x=x\)<=>\(x=0\)

d. \(B=\frac{x-2}{x+2}< 1\)<=>\(x-2< x+2\)luôn đúng \(\forall\)x\(\ne\pm2\)

e. \(B=\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=1-\frac{4}{x+2}\)

Để B nguyên thì 4/x+2 nguyên => x + 2\(\in\){ - 4 ; - 2 ; - 1 ; 1 ; 2 ; 4 }

=> x \(\in\){ - 6 ; - 4 ; - 3 ; - 1 ; 0 ; 2 }