Câu 9: Trang 24 sách VNEN 9 tập 1
Chứng minh đẳng thức:
a) \sqrt{6}$ + 2$\sqrt{\frac{2}{3}}$ - 4$\sqrt{\frac{3}{2}}$ = $\frac{\sqrt{6}}{6}$ ;
b) : $\frac{1}{\sqrt{x} - \sqrt{y}}$ = x - y với x > 0, y > 0, x $\neq $ y ;
c) + $\frac{\sqrt{x}}{y - \sqrt{xy}}$ = $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$ với x > 0, y > 0, x $\neq $ y.
Bài làm:
Giải câu a)
Ta có: \sqrt{6}$ + 2$\sqrt{\frac{2}{3}}$ - 4$\sqrt{\frac{3}{2}}$ = $\frac{3}{2}$.$\frac{6}{\sqrt{6}}$ + 2$\frac{\sqrt{2}.\sqrt{2}}{\sqrt{3}.\sqrt{2}}$- 4$\frac{\sqrt{3}.\sqrt{3}}{\sqrt{2}.\sqrt{3}}$ = $\frac{9}{\sqrt{6}}$ + $\frac{4}{\sqrt{6}}$ - $\frac{12}{\sqrt{6}}$ = $\frac{1}{\sqrt{6}}$ = $\frac{\sqrt{6}}{6}$ (đpcm)
Vậy \sqrt{6}$ + 2$\sqrt{\frac{2}{3}}$ - 4$\sqrt{\frac{3}{2}}$ = $\frac{\sqrt{6}}{6}$
Giải câu b)
Ta có: : $\frac{1}{\sqrt{x} - \sqrt{y}}$ = ().($\sqrt{x}$ - $\sqrt{y}$) = $\frac{\sqrt{xy}(\sqrt{x} + \sqrt{y})}{\sqrt{xy}}$.($\sqrt{x}$ - $\sqrt{y}$) = ($\sqrt{x}$ + $\sqrt{y}$)($\sqrt{x}$ - $\sqrt{y}$) = x - y (đpcm)
Vậy : $\frac{1}{\sqrt{x} - \sqrt{y}}$ = x - y.
Giải câu c)
Ta có: + $\frac{\sqrt{x}}{y - \sqrt{xy}}$ = $\frac{\sqrt{y}}{\sqrt{x}(\sqrt{x} - \sqrt{y})}$ + $\frac{\sqrt{x}}{\sqrt{y}(\sqrt{y} - \sqrt{x})}$ = $\frac{y}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ - $\frac{x}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ = - $\frac{x - y}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ = - $\frac{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ = - $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$ (đpcm)
Vậy + $\frac{\sqrt{x}}{y - \sqrt{xy}}$ = $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$.