a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b) Thay \(x=7-4\sqrt{3}\) ta có:
\(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\\ =\dfrac{\sqrt{7-4\sqrt{3}}-5}{\sqrt{7-4\sqrt{3}}+5}\\ =\dfrac{\sqrt{7-2\sqrt{12}}-5}{\sqrt{7-4\sqrt{12}}+5}\\ =\dfrac{\sqrt{7-2\sqrt{12}}-5}{\sqrt{7-4\sqrt{12}}+5}\)
bạn xem đúng đề không, đến đây mình ko giải tiếp dc
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
\(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: Thay \(x=7-4\sqrt{3}\) vào A, ta được:
\(A=\dfrac{2-\sqrt{3}-5}{2-\sqrt{3}+5}=\dfrac{-3-\sqrt{3}}{7-\sqrt{3}}=\dfrac{-12-5\sqrt{3}}{23}\)
