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f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
a/ \(\frac{\sqrt{a}-\left(\sqrt{a}\right)^2}{\sqrt{a}-1}\)
=\(\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{\sqrt{a}-1}\)
=\(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\)
=\(-\sqrt{a}\)
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
\(A=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)
\(A=\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)+y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{x\sqrt{xy}-x^2+y\sqrt{xy}+y^2-y^2+x^2}{\sqrt{xy}\left(y-x\right)}\)
\(A=\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}=\frac{y+x}{y-x}\)
KO CÓ GIÁ TRỊ y sao tính đây !!!!!!
CÒN \(x=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\) nhé