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23.阅读下列解题过程，解答问题.
$\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\sqrt{(\dfrac{1}{2})^2}=\dfrac{1}{2}$；$\sqrt{1-\dfrac{5}{9}}=\sqrt{\dfrac{4}{9}}=\sqrt{(\dfrac{2}{3})^2}=\dfrac{2}{3}$；
$\sqrt{1-\dfrac{7}{16}}=\sqrt{\dfrac{9}{16}}=\sqrt{(\dfrac{3}{4})^2}=\dfrac{3}{4}$；
……
$(1)\sqrt{1-\dfrac{9}{25}}=$，$\sqrt{1-\dfrac{15}{64}}=$；
(2)观察上面的解题过程，求$\sqrt{1-\dfrac{2n+1}{(n+1)^2}}（n$为自然数）；
(3)计算：
$\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\sqrt{(\dfrac{1}{2})^2}=\dfrac{1}{2}$；$\sqrt{1-\dfrac{5}{9}}=\sqrt{\dfrac{4}{9}}=\sqrt{(\dfrac{2}{3})^2}=\dfrac{2}{3}$；
$\sqrt{1-\dfrac{7}{16}}=\sqrt{\dfrac{9}{16}}=\sqrt{(\dfrac{3}{4})^2}=\dfrac{3}{4}$；
……
$(1)\sqrt{1-\dfrac{9}{25}}=$，$\sqrt{1-\dfrac{15}{64}}=$；
(2)观察上面的解题过程，求$\sqrt{1-\dfrac{2n+1}{(n+1)^2}}（n$为自然数）；
(3)计算：
答案
(1) $\dfrac{4}{5}$，$\dfrac{7}{8}$；(2) $\dfrac{n}{n+1}$；(3) $\dfrac{1}{50}$。
解析
解：
(1)
$\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\sqrt{(\dfrac{4}{5})^2}=\dfrac{4}{5}$，
$\sqrt{1-\dfrac{15}{64}}=\sqrt{\dfrac{49}{64}}=\sqrt{(\dfrac{7}{8})^2}=\dfrac{7}{8}$；
(2)
$\sqrt{1-\dfrac{2n+1}{(n+1)^2}}$
$=\sqrt{\dfrac{(n+1)^2-(2n+1)}{(n+1)^2}}$
$=\sqrt{\dfrac{n^2+2n+1-2n-1}{(n+1)^2}}$
$=\sqrt{\dfrac{n^2}{(n+1)^2}}$
$=\sqrt{(\dfrac{n}{n+1})^2}$
$=\dfrac{n}{n+1}$
(3)
原式$=\sqrt{(\dfrac{1}{2})^2}×\sqrt{(\dfrac{2}{3})^2}×\sqrt{(\dfrac{3}{4})^2}×···×\sqrt{(\dfrac{49}{50})^2}$
$=\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{4}×···×\dfrac{49}{50}$
$=\dfrac{1}{50}$
(1)
$\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\sqrt{(\dfrac{4}{5})^2}=\dfrac{4}{5}$，
$\sqrt{1-\dfrac{15}{64}}=\sqrt{\dfrac{49}{64}}=\sqrt{(\dfrac{7}{8})^2}=\dfrac{7}{8}$；
(2)
$\sqrt{1-\dfrac{2n+1}{(n+1)^2}}$
$=\sqrt{\dfrac{(n+1)^2-(2n+1)}{(n+1)^2}}$
$=\sqrt{\dfrac{n^2+2n+1-2n-1}{(n+1)^2}}$
$=\sqrt{\dfrac{n^2}{(n+1)^2}}$
$=\sqrt{(\dfrac{n}{n+1})^2}$
$=\dfrac{n}{n+1}$
(3)
原式$=\sqrt{(\dfrac{1}{2})^2}×\sqrt{(\dfrac{2}{3})^2}×\sqrt{(\dfrac{3}{4})^2}×···×\sqrt{(\dfrac{49}{50})^2}$
$=\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{4}×···×\dfrac{49}{50}$
$=\dfrac{1}{50}$
