Đáp án:
Giải thích các bước giải:
A = $\frac{7}{4}$.($\frac{3333}{1212}$ + $\frac{3333}{2020}$ + $\frac{3333}{3030}$ + $\frac{3333}{4242}$)
= $\frac{7}{4}$.($\frac{33}{12}$ + $\frac{33}{20}$ + $\frac{33}{30}$ + $\frac{33}{42}$)
= $\frac{7}{4}$.(33.$\frac{1}{12}$ + 33.$\frac{1}{20}$ + 33.$\frac{1}{30}$ + 33.$\frac{1}{42}$)
= $\frac{7}{4}$.(33.[$\frac{1}{12}$ + $\frac{1}{20}$ + $\frac{1}{30}$ + $\frac{1}{42}$])
= $\frac{7}{4}$.(33.[$\frac{1}{3.4}$ + $\frac{1}{4.5}$ + $\frac{1}{5.6}$ + $\frac{1}{6.7}$])
= $\frac{7}{4}$.(33.[$\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{6}$ + $\frac{1}{6}$ - $\frac{1}{7}$])
= $\frac{7}{4}$.(33.[$\frac{1}{3}$ - $\frac{1}{7}$])
= $\frac{7}{4}$.(33.[$\frac{7}{21}$ - $\frac{3}{21}$])
= $\frac{7}{4}$.(33.$\frac{4}{21}$)
= $\frac{7}{4}$.(11.$\frac{4}{7}$)
= $\frac{7}{4}$.$\frac{44}{7}$
= $\frac{1}{1}$.$\frac{11}{1}$
= 11
