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

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

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

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:

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

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

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

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

Bạn cộng các mẫu trong hoặc và giữ nguyên tử nếu kết quả trong hoặc rút gọn đc thì rút luôn. Đây là cách làm trong hoặc. Tính trong hoặc xong bạn chỉ việc nhân lại với nhau thôi, kết quả cuối cùng rút đc thì rút luôn( ko đc thì thôi, đừng cố rút gọn)

A=7/4.(11/4+33/20+11/10+11/14

A=7/4.(385/140+231/140+154/140+110/140)

A=7/4.(880/140)

A=7/4.44/7

A=11

k mình nhé

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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