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

\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)

\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)

\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)

\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)

\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)

A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)

A=7/4.(33/12+33/20+33/30+33/42)

A=7/4.33.(1/3*4+1/4*5+1/5*6+1/6*7)

A=231/4.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)

A=231/4.(1/3-1/7)

A=231/4.4/21

A=11. Vay A=11

Nho k cho minh voi nhe

A= 7/4-(33/12x101+33/20x101+33/30x101+33/42x101)

=7/4-[101x(33/12+33/20+33/30+33/42)]

=7/4-44/7

=-127/28

khăm thế đủ chưa

trả lời nè: A=\(\frac{1347}{202}\)(mk bấm máy tính là ra liền )

\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)

\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)

\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)

mk lm chak vớ vẩn rồi

\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)

\(A=\frac{7}{4}\left(\frac{385}{140}+\frac{231}{140}+\frac{154}{140}+\frac{110}{140}\right)\)

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

\(A=\frac{44}{4}=11\)

=> A=\(\frac{7}{4}\) . ( \(\frac{33}{12}\) + \(\frac{33}{20}\) + \(\frac{33}{30}\) + \(\frac{33}{42}\) ) => A=   \(\frac{7}{4}\).33. ( \(\frac{1}{12}\) + \(\frac{1}{20}\) + \(\frac{1}{30}\) + \(\frac{1}{42}\) )

=> A=\(\frac{7}{4}\).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. \(\frac{4}{21}\) = 11. Vậy A=11

Ta có:

\(\Rightarrow A=\frac{7}{4}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)

\(\Rightarrow A=\frac{7}{4}.\frac{44}{7}=11\)