Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
ĐKXĐ: \(\left[{}\begin{matrix}x< -2\\x\ge-\frac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\frac{8}{5\left(x+2\right)}-1+\sqrt{\frac{5x+2}{x+2}}-\frac{6}{5}=0\)
\(\Leftrightarrow\frac{-\left(5x+2\right)}{5\left(x+2\right)}+\sqrt{\frac{5x+2}{x+2}}-\frac{6}{5}=0\)
Đặt \(\sqrt{\frac{5x+2}{x+2}}=a\ge0\)
\(-\frac{1}{5}a^2+a-\frac{6}{5}=0\Leftrightarrow a^2-5a+6=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{5x+2}{x+2}=4\\\frac{5x+2}{x+2}=9\end{matrix}\right.\) \(\Rightarrow...\)