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\(=x+2\sqrt{xy}+y-9\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2-3^2\)
\(=\left(\sqrt{x}+\sqrt{y}-3\right)\left(\sqrt{x}+\sqrt{y}+3\right)\)
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a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
b: \(B=5x^2-7x\sqrt{y}+2y\)
\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)
\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\left(x-2\sqrt{xy}+y\right)}{x-y}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
=1
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d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
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a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+1\)
=2
c: \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
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\(xy-y\sqrt{x}+\sqrt{x}-1\)
\(=y\left(x-\sqrt{x}\right)+\left(\sqrt{x}-1\right)\)
\(=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)
\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)
b) Xét tử:
\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1)
Xét mẫu:
\(x+\sqrt{xy}+y\)
\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2)
Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm)
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