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a: \(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}=\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}\)
\(=\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}=1-\sqrt2\)
b: \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{1+\sqrt{xy}}\)
\(=\frac{x-2\sqrt{xy}+y+4\sqrt{xy}}{1+\sqrt{xy}}=\frac{x+2\sqrt{xy}+y}{1+\sqrt{xy}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{1+\sqrt{xy}}=\frac{\left(\sqrt2+\sqrt3\right)^2}{1+\sqrt{2\cdot3}}=\frac{5+2\sqrt6}{1+\sqrt6}\)
\(=\frac{\left(5+2\sqrt6\right)\left(\sqrt6-1\right)}{5}=\frac{5\sqrt6-5+12-2\sqrt6}{5}=\frac{3\sqrt6+7}{5}\)
a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)
b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)
c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)
\(y\ne0\)
Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)
e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)
Vì y < 0 nên \(\left|y\right|=-y\)
Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)
f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)
a)\(\left(2\sqrt{x}-3\right)\left(2+\sqrt{x}\right)+6=0\)
\(\Leftrightarrow4\sqrt{x}+2x-6-3\sqrt{x}+6=0\)
\(\Leftrightarrow2x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\2\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{4}\end{array}\right.\)