Vời mọi k > 0 , ta có:

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

\(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{50}}\)

\(\Rightarrow A< 2\left[\left(\sqrt{50}-\sqrt{49}\right)+...+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-0\right)\right]\)

\(\Rightarrow A< 2\left(\sqrt{50}-0\right)=2\sqrt{50}=10\sqrt{2}\)(đpcm)

thành 3 lần

1+1=2 thế mà cũng hỏi ?

So bi chia = 94

So chia = 29