\(P=\left[\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{y-x}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\left[\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\left[\sqrt{x}+\sqrt{y}-\frac{\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{x}\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)}:\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

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

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

\(=\frac{x+2\sqrt{xy}+y-\sqrt{xy}}{x-2\sqrt{xy}+y+\sqrt{xy}}\)

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

\(ĐKXĐ:\)

\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ne4\\x\ge0\end{cases}}\)

Vậy...................................................

\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)

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

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

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

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

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

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

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

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

chịu thua vô điều kiện xin lỗi nha : v

muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v

Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\Rightarrow A^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)

\(\Leftrightarrow A^3=\left(9+4\sqrt{5}\right)+\left(9-4\sqrt{5}\right)\)

                                      \(+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

Mà \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

Suy ra: \(A^3=18+3\sqrt[3]{9^2-\left(4\sqrt{5}\right)^2}.A\)

                   \(\Leftrightarrow A^3=18+3\sqrt[3]{81-80}.A\Leftrightarrow A^3=18+3A\Leftrightarrow A^3-3A-18=0\)

               \(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)

a) đặt A=\(\sqrt{2+\sqrt{3}}\)

=>    \(\sqrt{2}.A=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

=>     A = \(\frac{\sqrt{6}+\sqrt{2}}{2}\)

ý b là nhân thêm 2 vào r lm tương tự nha bn ! 

\(2x-5\sqrt{x}+2=0\)

\(\Leftrightarrow-5\sqrt{x}+2=0-2x\)

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

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

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

\(\Leftrightarrow25x=4x^2+8x+4\)

\(\Leftrightarrow4x^2+8x+4=25x\) (chuyển vế)

\(\Leftrightarrow4x^2+8x+4-25x=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

\(\Leftrightarrow x\left(4x-1\right)-\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=1\end{cases}}\)

Vậy nghiệm phương trình là: \(\left\{\frac{1}{4};1\right\}\)

ĐK:\(x\ge0\)

pt\(\Leftrightarrow\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Rightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{4}\\x=4\end{cases}}}\left(TMĐK\right)\)

Vậy....