Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a cong tru loan nen ko hieu
b
A=5/1.4+5/4.7+..5/100.103
3/5.A=3/1.4+3/4.7+..+3/100.103
=1/1-1/4+1/4-1/7+...+1/100-1/103
=1-1/103=102/103
A=(5.102)/(3.103)=5.34/103
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
Đặt biểu thức là A
=> \(A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
=> \(A=\frac{5}{2}.\frac{100}{101}\)
=> \(A=\frac{250}{101}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
Bài làm
D=ko viết lại đề
=1/1.3+1/1.5+1/5.7+1/7.9-1/2.4-1/4.6-1/6.8-1/8.10
=1+1/9-1-1/10
=10/9-9/10
=19/90
=(1/1.3+...+1/7.9)-(1/2.4+...+1/8.10)
=2(1/1.3+...+1/7.9)-2(1/2.4+...+1/8.10)
=(2/1.3+...+2/7.9)-(2/2.4+...+2/8.10)
=(1-1/3+...+1/7-1/9)-(1/2-1/4+ +1/8-1/10)
=1-1/9-1/2+1/10
tự tính tiếp nhé
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)
A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)
A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)
A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)
A = \(34+\frac{34}{103}=\frac{3536}{103}\)
bạn làm hộ mik câu B với