Giải câu 9 trang 34 toán VNEN 9 tập 1.
a) P = $\left ( \frac{\sqrt{x} - \sqrt{y}}{1 + \sqrt{xy}} + \frac{\sqrt{x} + \sqrt{y}}{1 - \sqrt{xy}} \right )$ : $\left ( \frac{x + y + 2xy}{1 - xy} + 1 \right )$
= $\left ( \frac{(\sqrt{x} - \sqrt{y})(1 - \sqrt{xy})}{(1 + \sqrt{xy})(1 - \sqrt{xy})} + \frac{(\sqrt{x} + \sqrt{y})(1 + \sqrt{xy})}{(1 - \sqrt{xy})(1 + \sqrt{xy})} \right )$ : $\frac{x + y + 2xy + 1 - xy}{1 - xy}$
= $\left ( \frac{\sqrt{x} - \sqrt{y} - x\sqrt{y} + y\sqrt{x}}{(1 + \sqrt{xy})(1 - \sqrt{xy})} + \frac{\sqrt{x} + \sqrt{y} + x\sqrt{y} + y\sqrt{x}}{(1 - \sqrt{xy})(1 + \sqrt{xy})} \right )$ : $\\frac{x + y + xy + 1}{1 - xy}$
= $\frac{\sqrt{x} - \sqrt{y} - x\sqrt{y} + y\sqrt{x} + \sqrt{x} + \sqrt{y} + x\sqrt{y} + y\sqrt{x}}{(1 + \sqrt{xy})(1 - \sqrt{xy})}$ : $\frac{x + y + xy + 1}{1 - xy}$
= $\frac{2\sqrt{x} + 2y\sqrt{x}}{1 - xy}$.$\frac{1 - xy}{(x + 1)(y+ 1)}$
= $\frac{2\sqrt{x}(1 + y)}{1 - xy}$.$\frac{1 - xy}{(x + 1)(y+ 1)}$
= $\frac{2\sqrt{x}}{x + 1}$.
b) Tại x = $\frac{2}{2 + \sqrt{3}}$ = $\frac{2(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}$ = $\frac{2(2 - \sqrt{3})}{4 - 3}$ = 4 - 2$\sqrt{3}$ = $(\sqrt{3} - 1)^{2}$
$\Rightarrow $ $\sqrt{x}$ = $\sqrt{3}$ - 1
Suy ra P = $\frac{2\sqrt{x}}{x + 1}$ = $\frac{2(\sqrt{3} - 1)}{4 - 2\sqrt{3} + 1}$ = $\frac{2 + 6\sqrt{3}}{13}$.
c) P = $\frac{2\sqrt{x}}{x + 1}$
Áp dụng bất đẳng thức Cô-si ta có:
x + 1 $\geq $ 2$\sqrt{x}$ $\Rightarrow $ P = $\frac{2\sqrt{x}}{x + 1}$ $\leq $ $\frac{2\sqrt{x}}{2\sqrt{x}}$ = 1. Dấu bằng khi x = 1
Vậy P $\leq $ 1.