Question

Chứng minh rằng:

3²/20.23+3²/23.26+...+3²/77.80<1

Answer 1

\(\dfrac{3^2}{20.23}\)+\(\dfrac{3^2}{23.26}\)+...+\(\dfrac{3^2}{77.80}\)

=> \(\dfrac{9}{20.23}+...+\dfrac{9}{77.80}\)

= 9.\(\left(\dfrac{1}{20.23}+...+\dfrac{1}{77.80}\right)\)

\(=9.\left(\dfrac{1}{20.3}-\dfrac{1}{23.3}+\dfrac{1}{23.3}-\dfrac{1}{26.3}+...+\dfrac{1}{77.3}-\dfrac{1}{80.3}\right)\)= \(9.\left(\dfrac{1}{20.3}-\dfrac{1}{80.3}\right)\)

\(=9.\dfrac{1}{80}\)=\(\dfrac{9}{80}=0,1125< 1.\)