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

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

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

mk lm chak vớ vẩn rồi

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

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

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

Ta có 

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

     \(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)

      \(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)

       \(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)

        \(=\frac{7}{4}.\left[\frac{3333}{101}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)

          \(=\frac{7}{4}.\frac{3333}{101}.\frac{4}{21}=\frac{1111}{101}\)

A= 11 nhé

Trả lời :A=11.

Đúng thì k cho mk nha.

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

\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(A=\frac{231}{4}.\frac{4}{21}=\frac{231}{21}=11\)

k nha

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

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

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

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

\(A=\frac{7}{4}\left[33\times\frac{4}{21}\right]\)

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

\(A=11\)

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

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

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

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

A= \(\frac{7}{4}.\left[33.\frac{4}{21}\right]\)

A= \(\frac{7}{4}.\frac{44}{7}\)

A= 11

                 Vậy A= 11

mik tính bằng máy ra \(\frac{363}{40}\)

ko chắc lắm hen

mình nghĩ là 11 vì dấu . ở lớp 6 là dấu nhân thì phải

A = 7/4 . (3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)
A = 7/4 . (11/4 + 33/20 + 11/10 + 11/14)
A = 7/4 . 44/7
A = 11
Chúc bạn học tốt

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

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

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

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

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

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

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

\(A=\frac{7}{4}.33.\frac{1}{6}\)

\(A=\frac{7.33}{4.6}\)

\(A=\frac{7.3.11}{4.3.2}\)

\(A=\frac{7.11}{4.2}\)

\(A=\frac{77}{8}\)

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

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

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\frac{4}{21}=11\)

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

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

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

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

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

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

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

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