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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: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
Ta có:\(\left(x+\sqrt{x^2+3}\right)\left(y-\sqrt{y^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(y-\sqrt{y^2+3}\right)\)
\(\Leftrightarrow-3\left(x+\sqrt{x^2+3}\right)=3\left(y-\sqrt{y^2+3}\right)\)
\(\Leftrightarrow x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\) (1)
Lại có:\(\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow-3\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\) (2)
Cộng theo vế \(\left(1\right)\) và \(\left(2\right)\) ta có:\(x+\sqrt{x^2+3}+y+\sqrt{y^2+3}=\sqrt{y^2+3}+\sqrt{x^2+3}-x-y\)
\(\Leftrightarrow2x+2y=0\Leftrightarrow x+y=0\)
Nhân cả 2 vế đẳng thức với \(\left(x-\sqrt{x^2+3}\right)\left(y-\sqrt{y^2+3}\right).\)
\(\Rightarrow VT=\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)\left(y-\sqrt{y^2+3}\right)=\)
\(=\left[x^2-\left(x^2+3\right)\right]\left[y^2-\left(y^2+3\right)\right]=\left(-3\right)\left(-3\right)=9\)
\(VP=3\left(x-\sqrt{x^2+3}\right)\left(y-\sqrt{y^2+3}\right)=VT=9\)
\(\Rightarrow\left(x-\sqrt{x^2+3}\right)\left(y-\sqrt{y^2+3}\right)=3=\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)\)
\(\Leftrightarrow xy-x\sqrt{y^2+3}-y\sqrt{x^2+3}+\sqrt{\left(x^2+3\right)\left(y^2+3\right)}=\)
\(=xy+x\sqrt{y^2+3}+y\sqrt{x^2+3}+\sqrt{\left(x^2+3\right)\left(y^2+3\right)}\)
\(\Leftrightarrow x\sqrt{y^2+3}=-y\sqrt{x^2+3}\)
\(\Leftrightarrow\sqrt{x^2\left(y^2+3\right)}=\sqrt{y^2\left(x^2+3\right)}\) Bình phương 2 vế
\(\Leftrightarrow x^2y^2+3x^2=x^2y^2+3y^2\Leftrightarrow x^2-y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=0\Rightarrow x+y=0\) với đk \(x\ne y\)