Tui nhầm đề xíu, cái A kia phải là:   A=\(\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)

thảo nào rút gọn mãi nó chả mất căn :))

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

\(=\sqrt{5}-1-\frac{5\sqrt{5}-10}{5}=\frac{5\sqrt{5}-5-5\sqrt{5}+10}{5}=\frac{5}{5}=1\)

Với \(x\ge0;x\ne4;9\)

\(P=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

\(=\left(\frac{3\sqrt{x}+6+\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\right):\left(\sqrt{x}+3\right)\)

\(=\left(\frac{x+5\sqrt{x}+6}{x-4}\right):\left(\sqrt{x}+3\right)=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(x-4\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-2}\)

b, \(2P-A< 0\Rightarrow\frac{2}{\sqrt{x}-2}-1< 0\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-2}< 0\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-2}>0\)

TH1 : \(\hept{\begin{cases}\sqrt{x}-4>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>16\\x>4\end{cases}\Leftrightarrow x>16}\)

TH2 : \(\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 16\\x< 4\end{cases}}\Leftrightarrow x< 4}\)

Kết hợp với đk vậy \(0\le x< 4;x>16\)

a. \(A=\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}=\left|1-\sqrt{5}\right|-\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}\)

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

 \(P=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)

\(=\frac{3\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

b. Ta có : 2P - A =\(\frac{2}{\sqrt{x}-2}-1=\frac{\sqrt{x}}{\sqrt{x}-2}\)

Để 2P - A < 0 thì \(\frac{x}{\sqrt{x}-2}< 0\)<=> \(\hept{\begin{cases}x< 0\\\sqrt{x}-2>0\end{cases}}\)hoặc\(\hept{\begin{cases}x>0\\\sqrt{x}-2< 0\end{cases}}\)

Mà x\(\ge\)0 nên \(\hept{\begin{cases}x>0\\\sqrt{x}-2< 0\end{cases}}\)<=>\(\hept{\begin{cases}x>0\\\sqrt{x}< 2\end{cases}}\)<=> 0 < x < 4 ( tmdk )

Vậy 2P - A = 0 <=> 0 < x < 4 

a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)

\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)

\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)

b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)

\(\Rightarrow9x+7\sqrt{x}-19=0\)

Mình k biết mình sai chỗ nào nữa, bạn xem giúp mình với

b/A=\(\frac{x-2\sqrt{x}-3-3\sqrt{x}+9}{x-2\sqrt{x}-3}=1-\frac{3\left(\sqrt{x}-3\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=1-\frac{3}{1+\sqrt{x}}\)

Vậy 1+ căn x thuốc Ư(3), mà \(\sqrt{x}\ge0\Rightarrow1+\sqrt{x}\ge1\)

Vậy \(1+\sqrt{x}=\left(1,3\right)\)

\(\Rightarrow\sqrt{x}=\left(0,2\right)\) Vì x nguyên nên x=0

\(\Leftrightarrow A=\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}:\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(\Leftrightarrow\frac{1}{1+\sqrt{x}}:\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{1+\sqrt{x}}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{x-9-x+4+\sqrt{x}+2}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{x-5\sqrt{x}+6}{x-2\sqrt{x}-3}\)

Lời giải:

a)

\(A=\left[\frac{4\sqrt{y}(\sqrt{y}-2)}{(2+\sqrt{y})(\sqrt{y}-2)}-\frac{8y}{(\sqrt{y}-2)(\sqrt{y}+2)}\right]:\left[\frac{\sqrt{y}-1}{\sqrt{y}(\sqrt{y}-2)}-\frac{2(\sqrt{y}-2)}{\sqrt{y}(\sqrt{y}-2)}\right]\)

\(=\frac{-4y-8\sqrt{y}}{(\sqrt{y}+2)(\sqrt{y}-2)}: \frac{3-\sqrt{y}}{\sqrt{y}(\sqrt{y}-2)}\)

\(=\frac{-4\sqrt{y}(\sqrt{y}+2)}{(\sqrt{y}+2)(\sqrt{y}-2)}.\frac{\sqrt{y}(\sqrt{y}-2)}{3-\sqrt{y}}\)

\(=\frac{4y}{\sqrt{y}-3}\)

b) Với \(y>0; y\neq 4; 9\):

Để \(A=2\Leftrightarrow \frac{4y}{\sqrt{y}-3}=2\Leftrightarrow 4y=2\sqrt{y}-6\)

\(\Leftrightarrow 4y-2\sqrt{y}=-6\)

\(\Leftrightarrow 3y+(\sqrt{y}-1)^2=-5< 0\) (vô lý với mọi $y>0$)

Do đó không tồn tại $y$ để $A=2$

x-9=(cănx-3)(cănx+3)

x+cănx-6=(cănx-2)(cănx+3)=-(2-cănx)(cănx+3)

x-3cănx=x(căn-3)

tự quy đồng rút gọn nha