Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)

b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

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

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

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

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

b) Sửa P thành A nha :b

Để \(A=\frac{-1}{3}\)

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

\(\Leftrightarrow2-\sqrt{x}=3\)

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

Vậy để \(A=-\frac{1}{3}\Leftrightarrow x\in\varnothing\)

Bạn tự tìm điều kiện xác định nhé :)

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

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

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

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

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

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

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

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

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

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

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

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)

ĐKXĐ: \(x>0\),\(x\ne3\)

\(P=\left(\frac{x-3\sqrt{x}}{x-6\sqrt{x}+9}-\frac{2\sqrt{x}-1}{x-3\sqrt{x}}\right).\frac{x-9}{\sqrt{x}+3}=\left[\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)^2}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}\right].\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=\left[\frac{x}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}\right].\left(\sqrt{x}-3\right)=\frac{x-2\sqrt{x}+1}{\sqrt{x}}\)

Bài làm:

Ta có: 

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

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

\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)

\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)