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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 \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)\)\(=\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=-1\)
\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right).\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2}.\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{2+\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
\(=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
\(=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
\(=-2\)
được bạn ạ mình nhờ thầy giải ra mà bạn tính máy tính mới ko ra thôi
\(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}=\left(3+\sqrt{3}\right)\sqrt{9-2.3.\sqrt{3}+3}\)
\(=\left(3+\sqrt{3}\right)\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left|3-\sqrt{3}\right|=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=6\)