Question

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

Answer 1

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

A=\(\frac{7}{4}.\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\) 

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

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

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

A=\(\frac{7}{4}.\frac{44}{7}\)

A=11

Like cho mình nha bài này viết mỏi tay lắm