Giải câu 9 trang 24 toán VNEN 9 tập 1

Giải câu a)

Ta có: $\frac{3}{2}$$\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 $\frac{3}{2}$$\sqrt{6}$ + 2$\sqrt{\frac{2}{3}}$ - 4$\sqrt{\frac{3}{2}}$ = $\frac{\sqrt{6}}{6}$

Giải câu b)

Ta có: $\frac{x\sqrt{y} + y\sqrt{x}}{\sqrt{xy}}$ : $\frac{1}{\sqrt{x} - \sqrt{y}}$ = ($\frac{x\sqrt{y} + y\sqrt{x}}{\sqrt{xy}}$).($\sqrt{x}$ - $\sqrt{y}$) = $\frac{\sqrt{xy}(\sqrt{x} + \sqrt{y})}{\sqrt{xy}}$.($\sqrt{x}$ - $\sqrt{y}$) = ($\sqrt{x}$ + $\sqrt{y}$)($\sqrt{x}$ - $\sqrt{y}$) = x - y (đpcm)

Vậy $\frac{x\sqrt{y} + y\sqrt{x}}{\sqrt{xy}}$ : $\frac{1}{\sqrt{x} - \sqrt{y}}$ = x - y.

Giải câu c)

Ta có: $\frac{\sqrt{y}}{x - \sqrt{xy}}$ + $\frac{\sqrt{x}}{y - \sqrt{xy}}$ = $\frac{\sqrt{y}}{\sqrt{x}(\sqrt{x} - \sqrt{y})}$ + $\frac{\sqrt{x}}{\sqrt{y}(\sqrt{y} - \sqrt{x})}$ = $\frac{y}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ - $\frac{x}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ = - $\frac{x - y}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ = - $\frac{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}{\sqrt{xy}(\sqrt{x} - \sqrt{y})}$ =  - $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$ (đpcm)

Vậy $\frac{\sqrt{y}}{x - \sqrt{xy}}$ + $\frac{\sqrt{x}}{y - \sqrt{xy}}$ = $\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$.