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 ! :)
A=\(3 \left(\right. \frac{3}{20.23} + \frac{3}{23.26} + \frac{3}{26.29} + . . . + \frac{3}{77.80} \left.\right)\)
A\(= 3 \left(\right. \frac{1}{20} - \frac{1}{23} + \frac{1}{23} - \frac{1}{26} + \frac{1}{26} - \frac{1}{29} + . . . + \frac{1}{77} - \frac{1}{80} \left.\right)\)
\(A = 3 \left(\right. \frac{1}{20} - \frac{1}{80} \left.\right)\)
A\(= 3 \left(\right. \frac{4}{80} - \frac{1}{80} \left.\right)\)
A\(= 3. \frac{3}{80}\)
A\(= \frac{9}{80}\)
\(\frac{9}{80} < 1\)
⇒A<1
