Question

Tính A= \(\frac{7}{4}\)x (\(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{3030}\))

Answer 1

A = \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{3030}\right)\)

A = \(\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

A = \(\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

A = \(\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

A = \(\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{7}{4}.33.\frac{4}{21}\)

=> A = \(\frac{1}{3}.33\)

=> A = 11