1) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)

Xin lỗi xin lỗi :v

1)\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)

= \(\sqrt{7}.\left(3\sqrt{7}-2\sqrt{14}\right)+14\sqrt{2}\)

= 21 - \(14\sqrt{2}+14\sqrt{2}\)

= 21

2) \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)

= \(\left(2\sqrt{2}-\sqrt{2}-\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{5}-2\sqrt{2}\right)\)

= \(\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)\)

=\(\left(\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2\)

= -3

\(\frac{\left(2\sqrt{8}+3\sqrt{3}+1\right)}{\sqrt{6}}\)

= \(\frac{4\sqrt{2}+3\sqrt{3}+1}{\sqrt{6}}\)

= \(\frac{8\sqrt{3}+9\sqrt{2}+6}{6}\)

\(\frac{\left(2\sqrt{8}+3\sqrt{3}+1\right)}{\sqrt{6}}=\frac{4\sqrt{2}+3\sqrt{3}+1}{\sqrt{6}}=\frac{8\sqrt{3}+9\sqrt{2}+6}{6}\)

~Study well~ :)

Điều kiện: \(n\ge1\)

\(B^2=\left(2\sqrt{n}\right)^2=4n\)

\(A^2=n-1+n+1+2\sqrt{\left(n-1\right)\left(n+1\right)}=2n+2\sqrt{n^2-1}\)

\(< 2n+2\sqrt{n^2}=2n+2\left|n\right|=2n+2n=4n=B^2\)

\(\Rightarrow A< B\)(vì A;B > 0)

\(\frac{\left(5\sqrt{7}+7\sqrt{5}\right)}{\sqrt{35}}\)

= \(\frac{\sqrt{5}.\left(\sqrt{35}+7\right)}{\sqrt{35}}\)

= \(\frac{\sqrt{35}+7}{\sqrt{7}}\)

= \(\sqrt{5}+\sqrt{7}\)

\(\frac{5\sqrt{7}+7\sqrt{5}}{\sqrt{35}}=\frac{\sqrt{5}.\sqrt{5}.\sqrt{7}+\sqrt{7}.\sqrt{7}.\sqrt{5}}{\sqrt{35}}.\)

\(=\frac{\sqrt{5}.\sqrt{35}+\sqrt{7}.\sqrt{35}}{\sqrt{35}}\)

\(=\frac{\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right)}{\sqrt{35}}=\sqrt{5}+\sqrt{7}\)