Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b. \(Q=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
c. Để \(Q< 1\Rightarrow Q-1< 0\Leftrightarrow\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Vậy \(0\le x< 9\)thì \(Q< 1\)
\(Q=\left(\frac{\sqrt{x}^2-1}{2\sqrt{x}}\right)^2.\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(Q=\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}\right].\left[\frac{\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(Q=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}.\frac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(Q=\frac{-4\sqrt{x}}{2\sqrt{x}}=-2\)
\(A=\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right):\left(\frac{1}{x-\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}-1}\right)\) Đkxđ : x > 1
\(A=\left(\frac{\sqrt{x}-1}{x-1}+\frac{2}{x-1}\right):\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(A=\frac{\sqrt{x}-1+2}{x-1}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{1-x}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(1-x\right)}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(1-x\right)}=\frac{\sqrt{x}}{1-x}\)
\(A=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left[\left(\frac{1}{2\sqrt{x}}\right)^2-2.\frac{1}{2\sqrt{x}}.\frac{\sqrt{x}}{2}+\left(\frac{\sqrt{x}}{2}\right)^2\right]\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right]\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)\)
\(\Leftrightarrow A=\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\left(\frac{1}{4x}-\frac{2x}{4x}+\frac{x^2}{4x}\right)\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{4\sqrt{x}}{1-x}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{1-x}{\sqrt{x}}\)
b) \(\frac{A}{\sqrt{x}}>1\)
\(\Leftrightarrow\frac{1-x}{\frac{\sqrt{x}}{\sqrt{x}}}>1\)
\(\Leftrightarrow1-x>1\Leftrightarrow x< 0\)