Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)
\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)
\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)
mk lm chak vớ vẩn rồi
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\frac{4}{21}=\frac{231}{21}=11\)
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=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
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
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\frac{1}{6}\)
\(A=\frac{7.33}{4.6}\)
\(A=\frac{7.3.11}{4.3.2}\)
\(A=\frac{7.11}{4.2}\)
\(A=\frac{77}{8}\)
\(A=\frac{7}{4}X\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}X\frac{44}{7}=11\)
\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}\left(\frac{385}{140}+\frac{231}{140}+\frac{154}{140}+\frac{110}{140}\right)\)
\(A=\frac{7}{4}.\frac{44}{7}\)
\(A=\frac{44}{4}=11\)
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