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17. （★★★）化简： $ \frac{3}{5}\sqrt{x y^{2}}÷(-\frac{4}{15}\sqrt{\frac{y}{x}}) $ $ (-\frac{5}{6}\sqrt{x^{3}y}). $
答案
17. 原式$=\frac{3}{5}\sqrt{xy^{2}}·(-\frac{15}{4}\sqrt{\frac{x}{y}})·(-\frac{5}{6}\sqrt{x^{3}y})$
$=(\frac{3}{5}×\frac{15}{4}×\frac{5}{6})·\sqrt{xy^{2}·\frac{x}{y}· x^{3}y}$
$=\frac{15}{8}\sqrt{x^{5}y^{2}}$
$=\frac{15}{8}x^{2}y\sqrt{x}$.
$=(\frac{3}{5}×\frac{15}{4}×\frac{5}{6})·\sqrt{xy^{2}·\frac{x}{y}· x^{3}y}$
$=\frac{15}{8}\sqrt{x^{5}y^{2}}$
$=\frac{15}{8}x^{2}y\sqrt{x}$.
18. （★★★）已知 $ a=\sqrt{\frac{2025}{2026}}, b=\sqrt{\frac{2026}{2027}}, $试比较 $ \frac{a}{b} $与1的大小.
答案
18. $\because a=\sqrt{\frac{2025}{2026}},b=\sqrt{\frac{2026}{2027}},$
$\therefore \frac{a}{b}=\frac{\sqrt{\frac{2025}{2026}}}{\sqrt{\frac{2026}{2027}}}=\sqrt{\frac{\frac{2025}{2026}}{\frac{2026}{2027}}}$
$=\sqrt{\frac{2025}{2026}×\frac{2027}{2026}}$
$=\sqrt{\frac{(2026 - 1)(2026 + 1)}{2026^{2}}}=\sqrt{\frac{2026^{2}-1}{2026^{2}}}$.
$\because 0＜\frac{2026^{2}-1}{2026^{2}}＜1,$
$\therefore \frac{a}{b}=\sqrt{\frac{2026^{2}-1}{2026^{2}}}＜1$,即$\frac{a}{b}＜1$.
$\therefore \frac{a}{b}=\frac{\sqrt{\frac{2025}{2026}}}{\sqrt{\frac{2026}{2027}}}=\sqrt{\frac{\frac{2025}{2026}}{\frac{2026}{2027}}}$
$=\sqrt{\frac{2025}{2026}×\frac{2027}{2026}}$
$=\sqrt{\frac{(2026 - 1)(2026 + 1)}{2026^{2}}}=\sqrt{\frac{2026^{2}-1}{2026^{2}}}$.
$\because 0＜\frac{2026^{2}-1}{2026^{2}}＜1,$
$\therefore \frac{a}{b}=\sqrt{\frac{2026^{2}-1}{2026^{2}}}＜1$,即$\frac{a}{b}＜1$.
