\(a,\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}=\sqrt{\left(3\sqrt{2}-2\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}\\ =3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\\ b,\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}\\ =\dfrac{\sqrt{n}-\sqrt{n+1}}{-1}=\sqrt{n+1}-\sqrt{n}\)

a) \(\sqrt{22-12\sqrt{2}}+\sqrt{6+4\sqrt{2}}\)

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

\(=3\sqrt{2}-2+2+\sqrt{2}=4\sqrt{2}\)

b) \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=\left(4+\sqrt{15}\right)\left(4-2\sqrt{15}\right).2\)

\(=\left(4^2-15\right).2\)

\(=2\left(ĐPCM\right)\)

\(VT=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+1-\sqrt{3}+1\)

=2=VP

\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)

\(=\sqrt{80}-\sqrt{2}\)

Đến đây bấm máy rồi đối chiếu kết quả cho nhanh, hoặc nếu em thik "màu mè" hơn thì giả sử lớn hơn rồi biến đổi tương đương thôi :)

Ta có \(\sqrt[4]{49+20\sqrt{6}}=\sqrt[4]{25+10\sqrt{24}+24}=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}\)

                               \(=\sqrt[4]{\left(\sqrt{3}+\sqrt{2}\right)^4}=\sqrt{3}+\sqrt{2}\)

Tương tự : \(\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\) ( Do \(\sqrt{3}>\sqrt{2}\) )

Suy ra \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

Lời giải:

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{79}+\sqrt{80}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+....+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(2A>\frac{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}{\sqrt{1}+\sqrt{2}}+\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{2}+\sqrt{3}}+....+\frac{(\sqrt{80}-\sqrt{79})(\sqrt{80}+\sqrt{79})}{\sqrt{79}+\sqrt{80}}+\frac{(\sqrt{81}-\sqrt{80})(\sqrt{81}+\sqrt{80})}{\sqrt{80}+\sqrt{81}}\)

\(2A>(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{80}-\sqrt{79})+(\sqrt{81}-\sqrt{80})\)

\(2A>\sqrt{81}-\sqrt{1}=8\)

\(A>4\) (đpcm)