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\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)
\(ĐK:x\ne\pm y\)
\(=\dfrac{2\left|x+y\right|}{2\left(x+y\right)\left(x-y\right)}=\dfrac{\sqrt{3}\left|x+y\right|}{\left(x+y\right)\left(x-y\right)}\)
Nếu x > -y thì x + y > 0 , ta có :\(\dfrac{\sqrt{3}}{x-y}\)
Nếu x < -y thì x + y < 0 , ta có :\(\dfrac{-\sqrt{3}}{x-y}\)
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 \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)
\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)
Sửa đề: \(A=\dfrac{1-3x}{2y}\cdot\sqrt{\dfrac{36y^2}{9x^2-6x+1}}\)
\(=\dfrac{1-3x}{2y}\cdot\sqrt{\left(\dfrac{6y}{3x-1}\right)^2}\)
\(=\dfrac{1-3x}{2y}\cdot\left|\dfrac{6y}{3x-1}\right|\)
x>1/3 nên 3x-1>0
y>0 nên 6y>0
=>\(A=\dfrac{1-3x}{2y}\cdot\dfrac{6y}{3x-1}=-3y\)
Có \(\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x+2xy+y\right)}{4}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\sqrt{\frac{3^2.\left(x+y\right)^2}{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}\frac{\sqrt{3^2}.\sqrt{\left(x+y\right)^2}}{\sqrt{2^2}}\)
=\(\frac{2}{\left(x+y\right)\left(x-y\right)}.\frac{3.\left(x+y\right)}{2}\)
=\(\frac{2.3.\left(x+y\right)}{\left(x+y\right)\left(x-y\right).2}\) =\(\frac{3}{x-y}\)
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
\(\dfrac{x+y}{y}.\sqrt{\dfrac{x^3y^2+2x^3y^2+xy^4}{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\sqrt{\dfrac{3x^3y^2+xy^4}{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{\sqrt{x^2+2xy+y^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{\sqrt{\left(x+y\right)^2}}\\ =\dfrac{x+y}{y}.\dfrac{\sqrt{3x^3y^2+xy^4}}{x+y}\\ =\dfrac{1}{y}.\sqrt{3x^3y^2+xy^4}\)