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 \(\frac{4x^2}{x^2-2x+1}\)

f) ĐKXĐ: \(x\ge-\frac{3}{2}\)

Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)

Lũy thừa 6 cả 2 vế lên PT tương đương:

\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)

Cái ngoặc to vô nghiệm vì nó tương đương:

\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))

Vậy x = 3.

PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra

@Akai Haruma, @Nguyễn Việt Lâm

giúp em vs ạ! Cần gấp ạ

em cảm ơn nhiều!

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

Đề có nhầm ko bạn?

\(E=2\Rightarrow\left(\frac{2x}{x-1}\right)^2=2\Rightarrow\left[{}\begin{matrix}\frac{2x}{x-1}=\sqrt{2}\\\frac{2x}{x-1}=-\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{2}x-\sqrt{2}\\2x=-\sqrt{2}x+\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-\sqrt{2}}{2-\sqrt{2}}< 0\left(l\right)\\x=\frac{\sqrt{2}}{2+\sqrt{2}}=\sqrt{2}-1\end{matrix}\right.\)