50+45=95,50/45+45/50=1

505 + 505 = 1010

50/45 + 45/50=2.01111111111 

\(505+505=1010\)

\(\frac{50}{45}+\frac{45}{50}=\frac{500}{450}+\frac{405}{450}=\frac{905}{450}=\frac{181}{90}\)

\(D=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)

\(D=\frac{2002.101}{101.12}+\frac{2002.101}{20.101}+\frac{2002.101}{30.101}+\frac{2002.101}{42.101}+\frac{2002.101}{56.101}\)

\(D=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)

\(D=\frac{1001}{6}+\frac{1001}{10}+\frac{1001}{15}+\frac{143}{3}+\frac{143}{4}\)

\(D=\frac{5005}{12}\)

đáp án là 5005/12 nhé bạn

tích cho mik nha

\(Rút\)\(gọn:\)\(\frac{1001}{6}+\frac{1001}{10}+\frac{1001}{15}+\frac{143}{3}+\frac{143}{4}\)

\(=\frac{1001}{3}+\frac{1001}{12}=\frac{5005}{12}\)

Ta có :\(\frac{230+x}{505+x}=\frac{4}{5}\)

\(\Rightarrow5.\left(230+x\right)=4.\left(505+x\right)\)

\(\Rightarrow1150+5x=2020+4x\)

\(\Rightarrow5x-4x=2020-1150\)

\(\Rightarrow x=870\)

Vậy x =870

X=870

X=870 đó bạn 

x = 28 / 87

Kết quả bằng 28 / 87

lời giải đàng hoàng chứ

\(\left(\frac{505}{707}+\frac{222}{333}\right)\)x=\(\frac{404}{909}\)

\(\left(\frac{5}{7}+\frac{2}{3}\right)\)x=\(\frac{4}{9}\)

\(\left(\frac{15}{21}+\frac{14}{21}\right)\)x=\(\frac{4}{9}\)

\(\frac{29}{21}\)x=\(\frac{4}{9}\)

x=\(\frac{4}{9}:\frac{29}{21}\)

x=\(\frac{4}{9}.\frac{21}{29}\)

x=\(\frac{28}{87}\)

Vậy x=\(\frac{28}{87}\)

k mình nha bạn

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

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

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

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

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

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

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

HD: Vũ Phương Vy  em chỉ cần đặt ts c rồi rút gọn

ko chép lại đề nha

=\(A=\frac{2\left(1-\frac{2}{19}+\frac{2}{23}\right)-\frac{1}{1010}}{3\left(1-\frac{1}{19}+\frac{1}{23}-\frac{1}{2020}\right)}\)\(.\frac{4\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{505}}{5\left(1-\frac{1}{29}+\frac{1}{41}\right)-\frac{1}{404}}\)

rồi em chỉ cần rút gọn tiếp 

p/s đến đây thấy đề kì kì sao đó 

em chek lại đề đc k

đề đúng rồi đó ạ

đề đúng mà ko sai đâu nếu bạn ko tin thì xem bài 5 của đề này nhé : https://online-learning-izteach-aws-source-bucket.s3-ap-southeast-1.amazonaws.com/resource/49714800-4b02-11ea-a775-01bd84d6799a/4baf4091-c5ae-4d0f-819c-e9ee9eedd6a8/de-so-13-on-tap-va-cung-co-toan-6-pdf

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)