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Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x;y>0\\x\ne y\end{cases}}\)
b) \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\right):\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\)
\(\Leftrightarrow A=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}:\frac{x+y}{y-x}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\frac{y-x}{x+y}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(y-x\right)}{x+y}\)
c) Thay \(x=4+2\sqrt{3},y=4-2\sqrt{3}\)vào A, ta được :
\(A=\frac{\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(4-2\sqrt{3}-4-2\sqrt{3}\right)}{4+2\sqrt{3}+4-2\sqrt{3}}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\right).\left(-4\sqrt{3}\right)}{8}\)
\(\Leftrightarrow A=\frac{\left(1+\sqrt{3}-\sqrt{3}+1\right).\left(-4\sqrt{3}\right)}{8}=\frac{-8\sqrt{3}}{8}=-\sqrt{3}\)
Vậy ....
Bài 1:
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}=\frac{2\sqrt{2\cdot4}-\sqrt{3\cdot4}}{\sqrt{2\cdot9}-\sqrt{16\cdot3}}-\frac{\sqrt{5}+\sqrt{9\cdot3}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}-\sqrt{2}}=\frac{\left(4\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)-\left(\sqrt{5}+3\sqrt{3}\right)\left(3\sqrt{2}-4\sqrt{3}\right)}{\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)}\)
\(=\frac{4\sqrt{60}-8-2\sqrt{90}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{3\sqrt{60}-6-4\sqrt{90}+4\sqrt{6}}\)
\(=\frac{8\sqrt{15}-8-6\sqrt{10}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{6\sqrt{15}-6-12\sqrt{10}+4\sqrt{6}}\)
\(=\frac{12\sqrt{15}-2\sqrt{10}-7\sqrt{6}+28}{6\sqrt{15}-12\sqrt{10}+4\sqrt{6}-6}\)
Đặt \(\hept{\begin{cases}\sqrt{5-x}=a\\\sqrt{x-3}=b\end{cases}}\)
=> a2 + b2 = 2
PT \(\Leftrightarrow\frac{a^3+b^3}{a+b}=2\Leftrightarrow\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a+b}=2\)
\(\Leftrightarrow2-ab=2\Leftrightarrow ab=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{5-x}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)