\(P=\sqrt{101-2\sqrt{101}+1}+\sqrt{101+2\sqrt{101}+1+1}\)

    \(=\sqrt{\left(\sqrt{101}-1\right)^2}+\sqrt{\left(\sqrt{101}+1\right)^2+1}>\sqrt{101}-1+\sqrt{101}+1=2\sqrt{101}>2.\sqrt{100}=2.10=20\)

=> P > 20

Áp dụng 

\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}<\frac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\)

có phải không?

trời ơi mk mà lm đc chắc đi thi hsg thế giới mất !!!

bạn nhật minh làm rồi mà 

Xét biểu thức phụ : 1 (2n+3)√2n+1+(2n+1)√2n+3 = 1 √2n+1.√2n+3(√2n+1+√2n+3) = √2n+3−√2n+1 √2n+1.√2n+3[(2n+3)−(2n+1)] = √2n+3−√2n+1 2√2n+1.√2n+3 = 1 2 ( 1 √2n+1 − 1 √2n+3 )với n≥1 Áp dụng : S= 1 3√1+1√3 + 1 3√5+5√3 + 1 5√7+7√5 +...+ 1 101√103+103√101 = 1 2 ( 1 √1 − 1 √3 )+ 1 2 ( 1 √3 − 1 √5 )+ 1 2 ( 1 √5 − 1 √7 )+...+ 1 2 ( 1 √101 − 1 √103 ) = 1 2 (1− 1 √3 + 1 √3 − 1 √5 + 1 √5 − 1 √7 +...+ 1 √101 − 1 √103 ) = 1 2 (1− 1 √103 )

Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)

Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)

DM CHƯA HỌC ĐẾN

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