A = 7/4 . ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242 07/10/2021 Bởi Aaliyah A = 7/4 . ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242
Đáp án: 11 Giải thích các bước giải: A = 7/4 . ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242) = 7/4 ( 33/12 + 33/20 + 33/30 + 33/42) =7/4 ( 33/3.4 + 33/4.5 + 33/5.6 + 33/6.7) =33. 7/4 ( 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7) = 33. 7/4 ( 1/3-1/4+1/4-1/5+1/5-…1/6-1/7) gạch ik số chùng = 33. 7/4 ( 1/3 + 1/7) = 33. 7/4 . 4/21 = 11 vậy A=11 Chúc bn hok tốt! Bình luận
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 Bình luận
Đáp án: 11
Giải thích các bước giải:
A = 7/4 . ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)
= 7/4 ( 33/12 + 33/20 + 33/30 + 33/42)
=7/4 ( 33/3.4 + 33/4.5 + 33/5.6 + 33/6.7)
=33. 7/4 ( 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7)
= 33. 7/4 ( 1/3-1/4+1/4-1/5+1/5-…1/6-1/7) gạch ik số chùng
= 33. 7/4 ( 1/3 + 1/7)
= 33. 7/4 . 4/21 = 11
vậy A=11
Chúc bn hok tốt!
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