[tex]A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\begin{bmatrix} \begin{pmatrix} \frac{1}{x}+\frac{1}{y} \end{pmatrix}.\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{(\sqrt{x}+\sqrt{y})^{3}}.\begin{pmatrix} \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} \end{pmatrix} \end{bmatrix}[/tex] với [tex]x=2-\sqrt{3}[/tex] và [tex]y=2+\sqrt{3}[/tex]. :r10bài này khó quá tớ nghĩ nát óc ko ra
ĐK: $x>0;y>0$
$A=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left [ \left ( \dfrac{1}{x}+\dfrac{1}{y} \right ).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac 2{(\sqrt x+\sqrt y)^3}.\left (\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}} \right ) \right ]
\\=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left [ \dfrac{x+y}{xy}.\dfrac{1}{(\sqrt{x}+\sqrt{y})^2}+\dfrac 2{(\sqrt x+\sqrt y)^3}.\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} \right ]
\\=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left [ \dfrac{x+y}{xy(\sqrt x+\sqrt y)^2}+\dfrac{2}{\sqrt{xy}(\sqrt x+\sqrt y)^2} \right ]
\\=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\dfrac{x+y+2\sqrt{xy}}{xy(\sqrt x+\sqrt y)^2}
\\=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}.\dfrac{xy(\sqrt x+\sqrt y)^2}{(\sqrt x+\sqrt y)^2}
\\=\dfrac{\sqrt x-\sqrt y}{\sqrt{xy}}$
Với $x=2-\sqrt 3;y=2+\sqrt 3$ (TMĐK) thì giá trị của bt $A$ là:
$A=\dfrac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}
\\\Rightarrow A\sqrt{2}=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}
\\=\sqrt{(\sqrt{3}-1)^2}-\sqrt{(\sqrt{3}+1)^2}
\\=\sqrt{3}-1-\sqrt{3}-1=-2
\\\Rightarrow A=-\sqrt 2$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
