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a) \(P=\dfrac{\left(x^2+2xy+9y^2\right)-\left(x+3y-2\sqrt{xy}\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x^2+6xy+9y^2\right)-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)^2-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)\left(x+3y-2\sqrt{xy}\right)}{x+3y-2\sqrt{xy}}\)
\(P=x+3y\)
b) \(\dfrac{P}{\sqrt{xy}+y}=\dfrac{x+3y}{\sqrt{xy}+y}=\dfrac{\left(x+3y\right):y}{\left(\sqrt{xy}+y\right):y}=\dfrac{\dfrac{x}{y}+3}{\sqrt{\dfrac{x}{y}}+1}\)
Đặt \(t=\sqrt{\dfrac{x}{y}}>0\) và \(\dfrac{P}{\sqrt{xy}+y}=Q\) thì \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)
\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)
Có \(\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x+2xy+y\right)}{4}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\sqrt{\frac{3^2.\left(x+y\right)^2}{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\frac{\sqrt{3^2}.\sqrt{\left(x+y\right)^2}}{\sqrt{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}.\frac{3.\left(x+y\right)}{2}\)
=\(\frac{2.3.\left(x+y\right)}{\left(x+y\right)\left(x-y\right).2}\) =\(\frac{3}{x-y}\)
\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)
b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{2\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}=\dfrac{6\sqrt{3}+2}{13}\)
\(\dfrac{x+y}{y}.\sqrt{\dfrac{x^3y^2+2x^3y^2+xy^4}{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\sqrt{\dfrac{3x^3y^2+xy^4}{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{\sqrt{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{\sqrt{\left(x+y\right)^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{x+y}\\ =\dfrac{1}{y}.\sqrt{3x^3y^2+xy^4}\)