\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)

\(=\sqrt{80}-\sqrt{2}\)

Đến đây bấm máy rồi đối chiếu kết quả cho nhanh, hoặc nếu em thik "màu mè" hơn thì giả sử lớn hơn rồi biến đổi tương đương thôi :)

Lời giải:

Gọi tổng trên là $A$. Ta có:

\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(2A>\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+...+\frac{\sqrt{81}-\sqrt{80}}{(\sqrt{80}+\sqrt{81})(\sqrt{81}-\sqrt{80})}\)

\(2A>(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{81}-\sqrt{80})\)

\(2A>\sqrt{81}-1=8\Rightarrow A>4\)

Ta có đpcm.

C/m: \(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k+1}-\sqrt{k}}{k+1-k}=\sqrt{k+1}-\sqrt{k}\)\(\left(k\ge1,k\in\text{ℕ}\right)\)

Có: \(\dfrac{1}{\sqrt{k-1}+\sqrt{k}}>\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\)

\(\Rightarrow\dfrac{2}{\sqrt{k-1}+\sqrt{k}}>\dfrac{1}{\sqrt{k}+\sqrt{k+1}}+\dfrac{1}{\sqrt{k-1}+\sqrt{k}}\)\(=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}=\sqrt{k+1}-\sqrt{k-1}\)

\(\Rightarrow2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\right)>\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{81}=9-1=8\)

\(\Rightarrow\dfrac{1}{\sqrt{1}+\sqrt{2}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}>4\)(đpcm).

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

Xét:

\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\)

\(\Rightarrow B=\sqrt{81}-\sqrt{1}=8\)

Mặt khác, do \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}< \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{2}{\sqrt{1}+\sqrt{2}}\)

Tương tự: \(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}< \frac{2}{\sqrt{3}+\sqrt{4}}\) ....

\(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}< \frac{2}{\sqrt{79}+\sqrt{80}}\)

Cộng vế với vế ta được: \(2A>B=8\Rightarrow A>4\)

Lời giải:

Ta thấy:

\(A-B=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{79}+\sqrt{80}}-\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

\(=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}-\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{77}+\sqrt{78}}>0\)

\(\Rightarrow A>B\)

Lời giải:

Đặt biểu thức đã cho là $A$

Ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{1+\sqrt{2}}> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\)

\(\Rightarrow \frac{1}{1+\sqrt{2}}> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)

Hoàn toàn TT: \(\frac{1}{\sqrt{3}+\sqrt{4}}> \frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)

.......

\(\frac{1}{\sqrt{79}+\sqrt{80}}> \frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

Cộng các bđt trên lại với nhau:

\(\Rightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

\(A> \frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\) (liên hợp)

\(A> \frac{1}{2}> (\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80})\)

\(A> \frac{1}{2}(\sqrt{81}-1)=4\) (đpcm)