a, ĐKXĐ : \(\left\{{}\begin{matrix}25-x\ne0\\x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x\ne25\\x\ge0\end{matrix}\right.\)
b, Ta có : \(B=\frac{3}{\sqrt{x}+5}-\frac{20-2\sqrt{x}}{25-x}\)
=> \(B=\frac{3}{\sqrt{x}+5}-\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(5-\sqrt{x}\right)}\)
=> \(B=\frac{3\left(5-\sqrt{x}\right)-20+2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(5-\sqrt{x}\right)}\)\(=\frac{-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(5-\sqrt{x}\right)}=-\frac{1}{5-\sqrt{x}}=\frac{1}{\sqrt{x}-5}\)
c, Ta có : \(\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\left|x-4\right|}{\sqrt{x}-5}\)
=> \(\left|x-4\right|=\sqrt{x}+2\)
=> \(\left[{}\begin{matrix}x-4=\sqrt{x}+2\\x-4=-\left(\sqrt{x}+2\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)=\sqrt{x}+2\\\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)=-\left(\sqrt{x}+2\right)\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\) ( TM )
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