a: ĐKXĐ: x-3>=0

=>x>=3

b: \(B=\sqrt2\left(\sqrt8-\sqrt3\right)+\sqrt6\)

\(=\sqrt{16}-\sqrt6+\sqrt6=\sqrt{16}\)

=4

c: \(C=\frac{x}{x+\sqrt{x}}+\frac{\sqrt{x}-1}{x-1}\)

\(=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}=\frac{2}{\sqrt{x}+1}\)

a) \(\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}.\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\)

b) \(\sqrt{\frac{3a}{5b}}=\frac{\sqrt{3a}}{\sqrt{5b}}=\frac{\sqrt{3a}.\sqrt{5b}}{5b}=\frac{\sqrt{15ab}}{5b}\left(a;b>0\right)\)

c) \(\sqrt{\frac{5}{12}}=\frac{\sqrt{5}}{\sqrt{12}}=\frac{\sqrt{5}.\sqrt{12}}{12}=\frac{\sqrt{60}}{12}=\frac{2\sqrt{15}}{12}=\frac{\sqrt{15}}{6}\)

d) \(\sqrt{\frac{5x}{18y}}=\frac{\sqrt{5x}}{\sqrt{18y}}=\frac{\sqrt{5x}}{\sqrt{3^2.2y}}=\frac{\sqrt{5x}}{3\sqrt{2y}}\)

\(=\frac{\sqrt{5x}.\sqrt{3y}}{3.2y}=\frac{\sqrt{15xy}}{6xy}\)

Quên mất k ghi đk xy > 0

a: \(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{x+y}{\sqrt{x}+\sqrt{y}}\)

Ta có: \(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{\left(x+y\right)_{}\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(\) \(=\frac{x^2-\sqrt{xy}\left(x+y\right)-y^2-x^2+y^2}{\sqrt{xy}\left(x-y\right)}=\frac{-\left(x+y\right)}{x-y}\)

b: Thay x=3; \(y=4+2\sqrt3\) vào A, ta được:

\(A=\frac{-\left(3+4+2\sqrt3\right)}{3-\left(4+2\sqrt3\right)}=\frac{-7-2\sqrt3}{-2\sqrt3-1}=\frac{7+2\sqrt3}{2\sqrt3+1}\)

\(=\frac{\left(7+2\sqrt3\right)\left(2\sqrt3-1\right)}{12-1}=\frac{14\sqrt3-7+12-2\sqrt3}{11}=\frac{12\sqrt3+5}{11}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\x\ne y\end{matrix}\right.\)

Gọi biểu thức trên là A , ta có:

\(A=\frac{2\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\frac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{3\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\frac{2\sqrt{x}-2\sqrt{y}+\sqrt{x}+\sqrt{y}-3\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\\ =\frac{-\sqrt{y}}{x-y}\left(=\frac{\sqrt{y}}{y-x}\right)\)

b) Với x=4 ; y=9 ta có:

\(A=\frac{\sqrt{9}}{9-4}=\frac{3}{5}\)

c) Ta có: với x>y>0 thì A<=>\(\left\{{}\begin{matrix}\sqrt{y}>0\\x>y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}>0\\y-x< 0\end{matrix}\right.\Leftrightarrow A< 0\)

Vậy A<0 với mọi x>y>0

\(A=\left(\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{y}-\sqrt{x}}\right):\dfrac{2\sqrt{xy}}{x-y}\)

\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)

b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)

Vậy pt vô nghiệm 

\(A=\left(5-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)

=5-xy