Tham khảo: Câu hỏi của Lương Tuấn Anh - Toán lớp 7 | Học trực tuyến

Ta có:

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)

...

\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).

Ta có :

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{`100}}=\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

........................................

\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+.......+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+........+\dfrac{1}{10}=\dfrac{100}{10}=10\)

\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+......+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right)\)

Giải:

Ta thấy:

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)

\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)

...................................

\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)

\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}.\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}.\)

\(>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}.\)
\(=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}\) (100 số hạng \(\dfrac{1}{10}\)).

\(=\dfrac{100}{10}=10.\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\left(đpcm\right).\)

Vậy..........

Ta có :

........................................

nhớ tìm kiếm trước khi hỏi

Ta có:

\(\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\)

\(\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\)

\(\sqrt{3}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{100}}\)

\(.............................\)

\(\sqrt{99}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\)

\(\sqrt{100}=\sqrt{100}\Rightarrow\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)

Cộng từng vế của các BĐT trên ta được:

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

\(=\dfrac{100}{\sqrt{100}}=\dfrac{100}{10}=10\)

Vậy \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>10\) (Đpcm)

VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)

\(=10-1=9=VP\)

\(\forall n\in N\) ta luôn có \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\) (*)

\(\Leftrightarrow\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)=1\)

\(\Leftrightarrow\left(n+1\right)-n=1\) (luôn đúng)

Vậy (*) được chứng minh.

Áp dụng với \(n=1;2;3;...;99\) ta có

\(S=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=\sqrt{100}-1=10-1=9\)

Vậy S là 1 số nguyên.

\(S=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\\ S=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\\ S=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\\ S=-1+\sqrt{100}=9\)

Ta có:

1/ căn 1> 1/10

1/ căn 2> 1/10

...

1/ căn 99> 1/10

1/ căn 100 = 1/10

=> 1/ căn 1 + 1/ căn 2 + ... + 1/ căn 99 + 1/ căn 100 > 100 . 1/10 = 10 (đpcm)

1/√1 > 1/10
1/√2 > 1/10
1/√3 > 1/10
....................
1/√99 > 1/10
1/√100 = 1/10
Cộng từng vế ta có:
1/√1 + 1/√2 + 1/√3 + ... + 1/√100 >100.1/0 = 10 (Đpcm)

\(\dfrac{1}{\sqrt{1}}\) > \(\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{2}}\)> \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}\) > \(\dfrac{1}{10}\)
....................
\(\dfrac{1}{\sqrt{99}}\) > \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}\) = \(\dfrac{1}{10}\)
Cộng theo vế ta có:
\(\dfrac{1}{\sqrt{1}}\)+\(\dfrac{1}{\sqrt{2}}\)+\(\dfrac{1}{\sqrt{3}}\)+ ... +\(\dfrac{1}{\sqrt{100}}\) >100.\(\dfrac{1}{10}\) = 10 (Đpcm)