Question

chứng minh rằng

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

Answer 1

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)

Answer 2

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)

Answer 3

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

\(\dfrac{1}{\sqrt{2}}\)> \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{3}}\) > \(\dfrac{1}{10}\)
....................
\(\dfrac{1}{\sqrt{99}}\) > \(\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}\) = \(\dfrac{1}{10}\)
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)