Giải câu 5 trang 23 toán VNEN 9 tập 1

a) Ta có:

$\sqrt{\frac{x}{y^{3}} + \frac{2x}{y^{4}}}$ = $\sqrt{\frac{xy + 2x}{y^{4}}}$ = $\frac{\sqrt{xy + 2x}}{y^{2}}$.

b) Ta có:

$\frac{x - \sqrt{xy}}{\sqrt{x} - \sqrt{y}}$ = $\frac{\sqrt{x}(\sqrt{x} - \sqrt{y})}{\sqrt{x} - \sqrt{y}}$= \sqrt{x}.

c) Ta có:

(a - b)$\sqrt{\frac{a^{2}b^{2}}{(a - b)^{2}}}$ = (a - b)$\sqrt{(\frac{ab}{(a - b))^{2}}}$ = (a - b).$\frac{ab}{a - b}$ = ab.

d) Ta có:

$\frac{a - \sqrt{3a} + 3}{a\sqrt{a} + 3\sqrt{3}}$ = $\frac{a - \sqrt{3a} + 3}{(\sqrt{a})^{3} + (\sqrt{3})^{3}}$ = $\frac{a - \sqrt{3a} + 3}{(\sqrt{a}+ \sqrt{3})((\sqrt{a})^{2} - \sqrt{3a} + (\sqrt{3})^{2})}$ = $\frac{a - \sqrt{3a} + 3}{(\sqrt{a}+ \sqrt{3})(a - \sqrt{3a} + 3)}$ = $\frac{1}{\sqrt{a}+ \sqrt{3}}$.