đặt 

\(A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\)

=>\(\sqrt{2}A=\sqrt{2}\sqrt{7+\sqrt{13}}+\sqrt{2}\sqrt{7-\sqrt{13}}\)

\(=\sqrt{14+2\sqrt{13}}+\sqrt{14-2\sqrt{13}}\)

\(=\sqrt{13+2\sqrt{13}+1}+\sqrt{13-2\sqrt{13}+1}\)

\(=\sqrt{\left(\sqrt{13}+1\right)^2}+\sqrt{\left(\sqrt{13}-1\right)^2}\)

\(=\sqrt{13}+1+\sqrt{13}-1=2\sqrt{13}\)

=>\(A=\frac{2\sqrt{13}}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}\sqrt{13}}{\sqrt{2}}=\sqrt{2}\sqrt{13}=\sqrt{26}\)

suy ra : ĐPCM

\(B=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{99}-\sqrt{97}}{99-97}.\)

\(=\frac{\sqrt{5}}{2}-\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2}-\frac{\sqrt{5}}{2}+...+\frac{\sqrt{99}}{2}-\frac{\sqrt{97}}{2}=\frac{\sqrt{99}}{2}-\frac{\sqrt{3}}{2}\)

Vậy \(B=\frac{\sqrt{99}-\sqrt{3}}{2}.\)

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Dùng kiến thức đồng dư là đơn giản nhất!

Xét mod 17

+3¹⁰ ≡ 8 =>3¹⁰⁰ = (3¹⁰)¹⁰ ≡ 8¹⁰ ≡ 13.

+5¹⁰ ≡ 9 => 5¹⁰⁰ = (5¹⁰)¹⁰ ≡ 9¹⁰ ≡ 13.

=> 3¹⁰⁰ + 5¹⁰⁰  ≡  13 + 13  ≡ 9.

Vậy số dư là 9.

Bài 2:

\(\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)