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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)
\(=\sqrt{80}-\sqrt{2}\)
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 :)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{79}+\sqrt{80}}\)
\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+....+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(2A>\frac{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}{\sqrt{1}+\sqrt{2}}+\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{2}+\sqrt{3}}+....+\frac{(\sqrt{80}-\sqrt{79})(\sqrt{80}+\sqrt{79})}{\sqrt{79}+\sqrt{80}}+\frac{(\sqrt{81}-\sqrt{80})(\sqrt{81}+\sqrt{80})}{\sqrt{80}+\sqrt{81}}\)
\(2A>(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{80}-\sqrt{79})+(\sqrt{81}-\sqrt{80})\)
\(2A>\sqrt{81}-\sqrt{1}=8\)
\(A>4\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(2A>B=8\Rightarrow A>4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta thấy:
\(A-B=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{79}+\sqrt{80}}-\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}-\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{77}+\sqrt{78}}>0\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)