Question

chứng minh \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}< 10\)

Answer 1

đề sai

chứng minh ngược lại C/m:>10

căn2<can3<can 4=>

1/căn2>1/căn3>1/căn4

1/căn2+1/can3+1/Căn4>3/can4=3/2

1/can5+....+1/can9>5.1/can9=5/3

1/can10+...+1/can16>7/can16=7/4

...

1/can81+...1/can100>18.1/can100= 19/10

A>B=1+3/2+5/3+7/4+...+19/10>10

Answer 2

Đề sai thật.

Xin phép sửa lại:

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

Giải:

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

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

....

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

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

Cộng từng vế trên HĐT ta có:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}\)

\(=10\)