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
Những câu hỏi liên quan
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:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
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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:
\(151B=\frac{51+100}{51.100}+\frac{52+99}{52.99}+....+\frac{99+52}{99.52}+\frac{100+51}{100.51}\)
\(=\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+....+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\)
\(=\left(\frac{1}{100}+\frac{1}{99}+....+\frac{1}{52}+\frac{1}{51}\right)+\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\right)=2A\)
\(\Rightarrow \frac{A}{B}=\frac{151}{2}\)
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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
\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)
\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)
\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)
\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)
\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)
\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)
\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)
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bạn giải thích kĩ đực không? Khó hiểu quá!
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Rút gọn B=1/51.100 + 1/52.99 +...+ 1/99.52 + 1/100.51
B=1/51.100+1/52.99+...+1/100.51
=>151B=1/51+1/100+1/52+1/99+...+1/100+1/51
=>151B/2=1/51+1/52+1/53+1/54+...+1/100
=>B=2/151.(1/51+1/52+1/53+1/54+...+1/100)
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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
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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~
Rút gọn: B=1/51.100+1/52.99+1/53.98+...+1/100.51
Giải thích nữa nhé!Cảm ơn trước
Tính tổng:
a, -1+3-5+7-....+97-99
b, 1+2-3-4+...+97+98-99-100
c, 1.2+2.3+3.4+4.5+....+99.100
Xem chi tiết
c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)
\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)
\(\Leftrightarrow A=33\cdot100\cdot101=333300\)
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b) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot25=-100\)
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A=(1/1•2+1/1•3+...+1/9•12).y = 99
A = 1/51 +1/52 + 1/53 + ... + 1/100
B = 1/1.2 + 1/3.4 + 1/4.5 + ... + 1/99.100
CMR A : B = 1
Làm mà t thấy hợp lí cho tick
Bài làm:
Dạ thưa đề B bạn viết sai rồi ạ!
Ta có: \(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{100}+\frac{1}{100}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{2}{2}+\frac{2}{4}+\frac{2}{6}+...+\frac{2}{100}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=A\)
\(A\div B=1\)
=> đpcm
Học tốt!!!!
ok tks bạn Đăng nhé <33
Cảm ơn bạn Tạ Duy Khoa nhìu ạ!
Cho B = \(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+ ... + \(\dfrac{1}{99.100}\)
Chứng minh \(\dfrac{7}{12}\)<B<\(\dfrac{5}{6}\)
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)
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