Question

Không dùng máy tính bỏ túi hãy so sánh :

A= \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}+\dfrac{1}{\sqrt{25}}\)và 5

ĐỀ THI TOÁN 9 TP BIÊN HÒA

Answer 1

Ủng hộ cách khác

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{25}};\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{25}};...;\dfrac{1}{\sqrt{24}}>\dfrac{1}{\sqrt{25}}\)

\(\Rightarrow A>\dfrac{1}{\sqrt{25}}+\dfrac{1}{\sqrt{25}}+...+\dfrac{1}{\sqrt{25}}=\dfrac{25}{\sqrt{25}}=5\)

Answer 2

\(A=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{25}}\)

\(A=2\left(\dfrac{1}{\sqrt{1}+\sqrt{1}}+\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{25}+\sqrt{25}}\right)\)

\(\Rightarrow A>2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{25}+\sqrt{26}}\right)\)

\(\Rightarrow A>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{26}-\sqrt{25}\right)\)

\(\Rightarrow A>2\left(\sqrt{26}-\sqrt{1}\right)>2\left(\sqrt{25}-\sqrt{1}\right)=8>5\)

Vậy \(A>5\) (thật ra lớn hơn hẳn 8 luôn, 5 chưa là gì cả :D)