Đề mắc lỗi hiển thị rồi. Bạn xem lại.

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?

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

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

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)

Với \(x\ge0\)

\(E=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)

\(=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\right).\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}\)

a) Ta có: \(E=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

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

\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để \(E=\dfrac{8}{9}\) thì \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}-36\sqrt{x}+24=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow24x-12\sqrt{x}-48\sqrt{x}+24=0\)

\(\Leftrightarrow12\sqrt{x}\left(2\sqrt{x}-1\right)-24\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(12\sqrt{x}-24\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\12\sqrt{x}-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}=1\\12\sqrt{x}=24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)

a)\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

c)Để E>0 thì \(\frac{x-1}{\sqrt{x}}>0\)

Mà \(\sqrt{x}>0\)

\(\Rightarrow x-1>0\) hay \(x>1\)

a: ĐKXĐ: x>0

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

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

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

b: E=2/5

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)

=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)

=>\(2x-3\sqrt{x}+4=0\)

=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)

=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)

=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)

Vậy: \(x\in\varnothing\)