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

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

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

\(=\frac{x+3\sqrt{x}-x+4\sqrt{x}-3}{\sqrt{x}+3}\)

\(=\frac{7\sqrt{x}-3}{\sqrt{x}+3}\)

\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x+4\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

Khó quá 

\(a,\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}-1+\sqrt{5}+1\)

\(=2\sqrt{5}\)

\(b,\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)(Đề sai sửa lại)

\(=\sqrt{5-2.2\sqrt{5}+4}+\sqrt{5+2.2\sqrt{5}+4}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\sqrt{5}-2+\sqrt{5}+2\)

\(=2\sqrt{5}\)

\(x-5\sqrt{x-2}=-2\)

\(\Leftrightarrow-5\sqrt{x-2}=-2-x\)

\(\Leftrightarrow\left(-5\sqrt{x-2}\right)^2=-2x-x\)

<=> 25x - 50 = 4 + 4x + x²

<=> x = 18 hoặc x = 3

Vậy:...

\(DKXĐ:x\ge2\)

\(x-5\sqrt{x-2}=-2\)

\(\Leftrightarrow5\sqrt{x-2}=x+2\)

\(\Leftrightarrow25\left(x-2\right)=\left(x+2\right)^2\)

\(\Leftrightarrow25x-50=x^2+4x+4\)

\(\Leftrightarrow x^2+4x+4-25x+50=0\)

\(\Leftrightarrow x^2-21x+54=0\)

\(\Leftrightarrow\left(x-18\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-18=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=18\\x=3\end{cases}\left(\frac{t}{m}ĐKXĐ\right)}\)

mình cần gấp, thanks các bạn

Đề chắc chắn đúng chứ bạn??

\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

\(\Rightarrow\sqrt{2x-3}=2\sqrt{x-1}\)

\(\Rightarrow2x-3=4\left(x-1\right)\)

\(\Rightarrow2x-3=4x-4\)

\(\Rightarrow4x-2x=4-3\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

\(ĐKXĐ:\hept{\begin{cases}x-1>0\\2x-3\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>1\\x>\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow x>\frac{3}{2}\)

Vậy nên \(x=\frac{1}{2}\) không thỏa mãn ĐKXĐ.

ta có 0<x<1<=>\(\sqrt{0}\)<\(\sqrt{x}\)<\(\sqrt{1}\)<=>0<\(\sqrt{x}\)<1           (1)

Nhân cả hai vế của bất đẳng thức \(\sqrt{x}\) <1 với \(\sqrt{x}\)ta được

   \(\sqrt{x}\).\(\sqrt{x}\)<1.\(\sqrt{x}\)

<=>            x  <\(\sqrt{x}\)

<=>     0   <\(\sqrt{x}\)-x

hay\(\sqrt{x}\)-x>0(đpcm)

Vậy...

KHÔNG BIẾT ĐÚNG KO , SAI THÔI NHA

Xét \(\sqrt{x}-x\) = \(-\left(x-\sqrt{x}\right)\)

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

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

 \(\left(\sqrt{x}-\frac{1}{2}\right)^2< \frac{1}{4}với.0< x< 1\)

\(\Rightarrow\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2>0\) với 0<x<1

hay \(\sqrt{x}-x>0\)với 0 <x<1

#mã mã#