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