Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
\(P=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}-x}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)
a) \(A=\left(1-\sqrt{18}+\sqrt{32}\right).\sqrt{3-2\sqrt{2}}\)
\(=\left(1-\sqrt{9.2}+\sqrt{16.2}\right).\sqrt{2-2\sqrt{2}+1}\)
\(=\left(1-\sqrt{9}.\sqrt{2}+\sqrt{16}.\sqrt{2}\right).\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(1-3\sqrt{2}+4\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
\(=\left(1+\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
Vì \(\sqrt{2}>1\)\(\Rightarrow\left|\sqrt{2}-1\right|>0\)
\(\Rightarrow A=\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)=\left(\sqrt{2}\right)^2-1=2-1=1\)
b) \(B=\frac{3}{6+\sqrt{35}}-\frac{3}{6-\sqrt{35}}=\frac{3\left(6-\sqrt{35}\right)}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}-\frac{3\left(6+\sqrt{35}\right)}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}\)
\(=\frac{18-3\sqrt{35}-18-3\sqrt{35}}{36-35}=-6\sqrt{35}\)