11.观察下列各式及验证过程:
$\sqrt{\frac{1}{2}-\frac{1}{3}}$=$\frac{1}{2}$$\sqrt{\frac{2}{3}}$,验证:$\sqrt{\frac{1}{2}-\frac{1}{3}}$=$\sqrt{\frac{1}{2×3}}$=$\sqrt{\frac{2}{{2}^{2}×3}}$=$\frac{1}{2}$$\sqrt{\frac{2}{3}}$;
$\sqrt{\frac{1}{2}(\frac{1}{3}-\frac{1}{4})}$=$\frac{1}{3}$$\sqrt{\frac{3}{8}}$,验证:$\sqrt{\frac{1}{2}(\frac{1}{3}-\frac{1}{4})}$=$\sqrt{\frac{1}{2×3×4}}$=$\sqrt{\frac{3}{2×{3}^{2}×4}}$=$\frac{1}{3}$$\sqrt{\frac{3}{8}}$;
$\sqrt{\frac{1}{3}(\frac{1}{4}-\frac{1}{5})}$=$\frac{1}{4}$$\sqrt{\frac{4}{15}}$,验证:$\sqrt{\frac{1}{3}(\frac{1}{4}-\frac{1}{5})}$=$\sqrt{\frac{1}{3×4×5}}$=$\sqrt{\frac{4}{3×{4}^{2}×5}}$=$\frac{1}{4}$$\sqrt{\frac{4}{15}}$;
$\sqrt{\frac{1}{4}(\frac{1}{5}-\frac{1}{6})}$=$\frac{1}{5}$$\sqrt{\frac{5}{24}}$,验证:$\sqrt{\frac{1}{4}(\frac{1}{5}-\frac{1}{6})}$=$\sqrt{\frac{1}{4×5×6}}$=$\sqrt{\frac{5}{4×{5}^{2}×6}}$=$\frac{1}{5}$$\sqrt{\frac{5}{24}}$;
(1)按照上述四个等式及其验证过程的基本思路,猜想$\sqrt{\frac{1}{5}(\frac{1}{6}-\frac{1}{7})}$的变形结果并进行验证;
(2)针对上述各式反映的规律,写出用n(n≥1为整数)表示的等式.