$解:(1)由前三個式子可知第4個等式為\sqrt{\frac{1}{4}×(\frac{1}{5}-\frac {1}{6})}=\frac{1}{5}\sqrt {\frac {5}{24}}$
$驗證: \sqrt{\frac{1}{4}×(\frac{1}{5}-\frac {1}{6})}=\sqrt{\frac{1}{4×5×6}}=\frac{1}{5}\sqrt {\frac {5}{24}}$
$\ 第 5 個等式為\sqrt{\frac{1}{5}×(\frac{1}{6}-\frac {1}{7})}=\frac{1}{6}\sqrt{\frac{6}{35}}$
$(2)由各式反映的規(guī)律可得\sqrt {\frac{1}{n}(\frac{1}{n+1}-\frac {1}{n+2})}=\frac{1}{n+1}\sqrt{\frac{n+1}{n(n+2)}}$
$驗證: \sqrt {\frac{1}{n}(\frac{1}{n+1}-\frac{1}{n+2}}=\sqrt{\frac{n+1}{n(n+1)^{2}(n+2)}}=\frac{1}{n+1}\sqrt{\frac{n+1}{n(n+2)}}$
