a: ĐKXĐ: x>=0; x<>25; x<>9
Sửa đề: \(J=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}-3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
Ta có: \(\frac{x-5\sqrt{x}}{x-25}-1\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\)
\(=\frac{\sqrt{x}}{\sqrt{x}+5}-1=\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}=\frac{-5}{\sqrt{x}+5}\)
Ta có: \(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}-3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\)
\(=\frac{\left(5+\sqrt{x}\right)\left(5-\sqrt{x}\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\)
\(=\frac{5-\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}-3}{\sqrt{x}+5}=\frac{-\sqrt{x}+3}{\sqrt{x}+5}\)
Ta có: \(J=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}-3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(\) \(=\frac{-5}{\sqrt{x}+5}:\frac{-\sqrt{x}+3}{\sqrt{x}+5}=\frac{5}{\sqrt{x}-3}\)
b: J<1
=>J-1<0
=>\(\frac{5-\sqrt{x}+3}{\sqrt{x}-3}<0\)
=>\(\frac{-\sqrt{x}+8}{\sqrt{x}-3}<0\)
=>\(\frac{\sqrt{x}-8}{\sqrt{x}-3}>0\)
=>\(\left[\begin{array}{l}\sqrt{x}-8>0\\ \sqrt{x}-3<0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}>8\\ \sqrt{x}<3\end{array}\right.\Rightarrow\left[\begin{array}{l}x>64\\ 0\le x<9\end{array}\right.\)
