Question

Bài 3 : Chứng minh rằng :

\(\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+...+\dfrac{3^2}{77.80}< 1\)

Answer 1

Đặt A= ...(như trên)

=>\(\dfrac{1}{3}A=\dfrac{1}{3}.\left(\dfrac{3^2}{20.23}+\dfrac{3^2}{23.26}+...+\dfrac{3^2}{77.80}\right)\)

=>\(\dfrac{1}{3}A=\dfrac{3}{20.23}+\dfrac{3}{23.26}+...+\dfrac{3}{77.80}\)

=>\(\dfrac{1}{3}A=\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+...+\dfrac{1}{77}-\dfrac{1}{80}\\ \)

=>\(\dfrac{1}{3}A=\dfrac{1}{20}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{4}{80}-\dfrac{1}{80}\\ =>\dfrac{1}{3}A=\dfrac{3}{80}=>A=\dfrac{3}{80}:\dfrac{1}{3}\\ =>A=\dfrac{3}{80}.3=\dfrac{9}{80}< 1\)

Vậy A<1 . Chúc bạn học tốt ! :)