ĐK: \(x\ge0,x\ne1\).

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

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

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

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

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

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

Với \(x=\frac{9}{4}\): 

\(A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=3\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}< 1\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-1}-1< 0\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow0\le x< 1\).

????????

Weo
-tt-

\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

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

\(=\frac{\sqrt{xy}.x+xy-xy-\sqrt{xy}.y}{x-y}\)

\(=\frac{\sqrt{xy}.x-\sqrt{xy}.y}{x-y}\)

\(\frac{=\sqrt{xy}.\left(x-y\right)}{x-y}\)

\(=\sqrt{xy}\)

a, \(\sqrt{16x}=8\)ĐK : x >= 0 

\(\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)ĐK : x >= 1 

\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)

t học lớp 8 bạn ei :)

ĐKXĐ : x \(\ge2\)

Ta có \(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)

\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}=2\sqrt{\left(x-2\right)\left(x+2\right)}-\left(x-2\right)-\left(x+2\right)+2\)

<=> \(\sqrt{x-2}-\sqrt{x+2}=-\left(\sqrt{x-2}-\sqrt{x+2}\right)^2+2\)

Đặt \(\sqrt{x-2}-\sqrt{x+2}=y\)

=> y = -y² + 2 

<=> y² - y - 2 = 0

<=> (y + 1)(y - 2) = 0

<=> \(\orbr{\begin{cases}y=-1\\y=2\end{cases}}\)

Khi y = -1

<=> \(\sqrt{x-2}-\sqrt{x+2}=-1\)

=> \(\left(\sqrt{x-2}-\sqrt{x+2}\right)^2=1\)

<=> \(\left(x-2\right)+\left(x+2\right)-2\sqrt{\left(x-2\right)\left(x+2\right)}=1\)

<=> \(2x-1=2\sqrt{\left(x-2\right)\left(x+2\right)}\)

=> 4x² - 4x + 1 = 2(x - 2)(x + 2) 

<=> 4x² - 4x + 1 = 2x² - 8 

<=> 2x² - 4x + 9 = 0 (vô lý) => TH1 loại

Khi y = 2 =>\(\sqrt{x-2}-\sqrt{x+2}=2\)

=> \(\left(\sqrt{x-2}-\sqrt{x+2}\right)^2=4\)

<=> \(2x-2\sqrt{\left(x-2\right)\left(x+2\right)}=4\)

<=> \(2x-4=2\sqrt{\left(x-2\right)\left(x+2\right)}\)

=> (2x - 4)² = 4(x - 2)(x + 2)

<=> 4(x - 2)² = 4(x - 2)(x + 2) 

<=> -16(x - 2) = 0

<=> x = 2 (tm) 

Vậy x = 2