Bài giải
1) $A=\frac{1}{\sqrt{\frac{5}{7}}+\sqrt{\frac{5}{13}}+1}+\frac{1}{\sqrt{\frac{7}{13}}+\sqrt{\frac{7}{5}}+1}+\frac{1}{\sqrt{1\frac{6}{7}}+\sqrt{2\frac{3}{5}}+1}\\
A=\frac{1}{\sqrt{\frac{5}{7}}+\sqrt{\frac{5}{13}}+\sqrt{\frac{5}{5}}}+\frac{1}{\sqrt{\frac{7}{13}}+ \sqrt{ \frac{7}{5}}+\sqrt{\frac{7}{7}}}+\frac{1}{ \sqrt{\frac{13}{7}}+\sqrt{\frac{13}{5}}+\sqrt{ \frac{13}{13}}}\\
A=\frac{1}{\sqrt{5}(\frac{1}{\sqrt{7}}+\frac{1}{ \sqrt{13}}+\frac{1}{\sqrt{5}})}+\frac{1}{\sqrt{7}( \frac{1}{\sqrt{13}}+\frac{1}{\sqrt{5}}+\frac{1}{ \sqrt{7}})}+\frac{1}{\sqrt{3}(\frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{7}}+\frac{1}{\sqrt{5}})}\\
A=\frac{1}{\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{13}}+\frac{1}{\sqrt{5}}}(\frac{1}{\sqrt{5}}+\frac{1}{ \sqrt{7}}+\frac{1}{\sqrt{13}})=1$
2) $\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4$
ĐK: x ≥ 11
$\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^{2}-x-11}=16\\
\Leftrightarrow x+\sqrt{x^{2}-x-11}=8\\
\Leftrightarrow \sqrt{x^{2}-x-11}=8-x$
Thêm ĐK: x ≤ 8
Mà x ≥ 11
$\to$ PT vô nghiệm
$\Rightarrow$ S = ∅