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?

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

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

Chào em, em có thể kam khảo tại link:

Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath

Nếu link bị chặn em copy và dán tại:

https://olm.vn/hoi-dap/question/1261852.html

Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath

a) Rút gọn E

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

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}-\left(\sqrt{x}-1\right)}\right]\)

\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)

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

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

\(E=\frac{x}{\sqrt{x}-1}\)

Vậy \(E=\frac{x}{\sqrt{x}-1}\)

bạn ơi ko có phần b c sao 

\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)

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

\(E=\left(\frac{\sqrt{x}-1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(E=\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=2\sqrt{x}\)

Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)

\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)

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 \(\frac{4x^2}{x^2-2x+1}\)