Ta co:\(\frac{1}{2^2}=\frac{1}{4}<\frac{1}{1.2}......\frac{1}{10}^2=\frac{1}{10.10}\)\(\)
Ta có:
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{9^2}<\frac{1}{8.9}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}<1\left(đpcm\right)\)
k mk nha
Dặt A là biểu thức trên
A<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
A<1-1/2+1/2-1/3+...+1/9-1/10
A<1-1/10<1
Vậy A<1
Mk nhầm
Ta có:
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{9^2}<\frac{1}{8.9}\)\(;\frac{1}{10^2}<\frac{1}{9.10}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)\(+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)\(+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}<1\left(đpcm\right)\)
k mk nha