Question

So sánh A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

với B=\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)

Answer 1

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)

\(=\sqrt{121}-\sqrt{1}=11-1=10\)

Lại có: \(\dfrac{1}{\sqrt{k}}=\dfrac{2}{2\sqrt{k}}>\dfrac{2}{\sqrt{k+1}+\sqrt{k}}\left(k>1\right)\)

\(\Leftrightarrow\dfrac{1}{\sqrt{k}}>\dfrac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{k+1-k}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

Áp dụng đánh giá trên vào B ta có:

\(B>1+2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\right)\)

\(=1+2\left(\sqrt{36}-\sqrt{2}\right)>1+2\left(6-1\right)=10\)

Suy ra \(A=10< B\Rightarrow A< B\)