a) Ta có: \(P=\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)
\(=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\right):\left(\frac{1-xy+x+y+2xy}{1-xy}\right)\)
\(=\left(\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}\right):\frac{x+y+xy+1}{1-xy}\)
\(=\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\frac{1-xy}{\left(x+1\right)+y\left(x+1\right)}\)
\(=\frac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}\)
\(=\frac{2\sqrt{x}}{x+1}\)
a, Đk: x ≥ 0, y ≥ 0, xy ≠ 1
P = \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{1-xy}\)
\(=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\frac{1+x+y+xy}{1-xy}\)
\(=\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}:\frac{\left(1+x\right)+y\left(1+x\right)}{1-xy}\)
\(=\frac{2\sqrt{x}\left(1+y\right)}{1-xy}:\frac{\left(1+x\right)\left(1+y\right)}{1-xy}\)
\(=\frac{2\sqrt{x}\left(1+y\right)}{1-xy}.\frac{1-xy}{\left(1+x\right)\left(1+y\right)}\)
\(=\frac{2\sqrt{x}}{1+x}\)
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b, Với x ≥ 0 ta có:
(\(\sqrt{x}\) - 1)2 ≥ 0
⇔ x - \(2\sqrt{x}\) + 1 ≥ 0
⇔ x + 1 ≥ \(2\sqrt{x}\)
⇔ \(\frac{x+1}{x+1}\ge\frac{2\sqrt{x}}{x+1}\)
⇔ \(1\ge\frac{2\sqrt{x}}{x+1}\)
⇔ \(\frac{2\sqrt{x}}{x+1}\) ≤ 1
Vậy Max P = 1 khi x = 1
