Rút gọn B=1/51.100 + 1/52.99 +...+ 1/99.52 + 1/100.51
Những câu hỏi liên quan
cho A=1/1.2+1/3.4+1/4.5+...+1/99.100
B=1/51.100+1/52.99+...+1/99.52+1/100.51
tính A/B
Cho A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100
B = 1/51.100 + 1/52.99 + ... + 1/99.52 + 1/100.51
Tính: A/B
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Mặt khác:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
Đúng 0
Bình luận (0)
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Mặt khác:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
Đúng 1
Bình luận (0)
cho C = 1/1.2 + 1/3.4 + 1/5.6 + . . . + 1/97.98 + 1/99.100 & D = 1/51.100 + 1/52.99 + 1/53.98 + . . . + 1/99.52 + 1/100.51
Chung minh C : D ko nhan gia tri la mot so tu nhien
\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)
\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)
\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)
\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)
\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)
\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)
\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)
Đúng 0
Bình luận (0)
bạn giải thích kĩ đực không? Khó hiểu quá!
Đúng 0
Bình luận (0)
Rút gọn: B=1/51.100+1/52.99+1/53.98+...+1/100.51
Giải thích nữa nhé!Cảm ơn trước
cho C=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
D=\(\frac{1}{51.100}\)+\(\frac{1}{52.99}\)+...+\(\frac{1}{99.52}\)+...+\(\frac{1}{100.51}\)
CMR : C:D là một số tự nhiên
Câu hỏi của Wang Jum Kai - Toán lớp 6 - Học toán với OnlineMath
Đúng 0
Bình luận (0)
Cho A=1/1.2+1/3.4+...+1/99.100
B=1/51.100+1/52.99+....+1/99.52+1/100.52
a) Rút gọn A và B
b) Tính A/B
Giải ra rõ ràng mình tick cho nha~
Mình là girl và là A.R.M.Y nha~
a) so sánh 5^567 và 3^852
b) cho A = 1-1/2+1/3-1/4+...+1/99-1/100 và 1/51.100 + 1/52.99 + ... + 1/75.76
tính tỉ số A/151B
rút gọn B = ( 1 - 1/2 ) x ( 1 -1/3 ) x ( 1 - 1/4 ) .... ( 1 -1/20 )
B= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).....\left(1-\dfrac{1}{20}\right)\)
B= \(\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{19}{20}\)
B= \(\dfrac{1.2.....19}{2.3.....20}\)
B= \(\dfrac{1}{20}\)
Đúng 0
Bình luận (0)
rút gọn B=1/2^2+1/2^3+....+1/2^100
B=1/22+1/23+...+1/2100
2B=1/21+1/22+...1/299
2B-B=(1/21+1/22+...+1/299)-(1/22+1/23+...+1/2100)
B=1/21-1/2100=299/2100-1/2100=299-1/2100
Đúng 0
Bình luận (0)