3x3=9 vì vậy bài toán trên sẽ chuyen thanh

50c\s 9 x50= ......( tự tính nghen)

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

\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)

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

\(A=\frac{7.11}{4}.\left(\frac{1}{4}+\frac{3}{20}+\frac{1}{10}+\frac{1}{14}\right)\)

\(A=\frac{77}{4}.\left(\frac{35}{140}+\frac{21}{140}+\frac{14}{140}+\frac{10}{140}\right)\)

\(A=\frac{77}{4}.\frac{80}{140}\)\(=\frac{77}{8}.\frac{20}{35}\)

\(A=11\)

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

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

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

\(A=\frac{7}{4}\cdot\left(33\cdot\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)

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

\(A=\frac{7}{4}\cdot\left(33\cdot\frac{4}{21}\right)\)

\(A=\frac{7}{4}\cdot\frac{132}{21}=11\)

ko hieu tai sao co the dat ra mot cau hoi vo li nhu vay?

deo hieuluon

B=333....3333  .    333....3333

B=\(333...3333^2\)

bang 3333o

k minh nhe

1111+2222+3333+1111+2222+3333+1111+2222+3333+1111+2222+3333+1111+2222+3333=(1111+2222+3333)x5

                                                                                                                                =33330  

- Vc Lớp 5 

32,2857 và 12,8 và -498814315,3 và 0,1 và 11 

tổng = - 498814259,1

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

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

\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)

\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{7}{4}\cdot33\cdot\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{7}{4}\cdot33\cdot\frac{4}{21}\)

\(=\left(\frac{7}{4}\cdot\frac{4}{21}\right)\cdot33\)

\(=\frac{1}{3}\cdot33=11\)