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KY
0
TN
2
YN
26 tháng 12 2021
Answer:
Mình làm thành tính tỉ số luôn nhé!
\(A=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}}\)
Ta xét \(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-1-\frac{1}{2}-...-\frac{1}{50}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}\)
\(\Rightarrow\frac{A}{B}=1\)
TN
8
19 tháng 1 2016
Xét mẫu số: 1/(2x3) + 1/(3x4) + …… + 1/(99x100)
= 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100
= (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)
= (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2
= (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2 + 1/3 + ....... +1/50 )
= 1/51 + 1/52 + 1/53 + ............. + 1/100 (Đơn giản số trừ)
Vậy: (1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/1x2 + 1/3x4 + .......... + 1/99x100) =
(1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1
NL
2
13 tháng 2 2016
B=1/1.2+1/3.4+1/5.6+...+1/99.100
=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100
=(1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+...+1/100)
=(1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100)-2(1/2+1/4+1/6+...+1/100)
=(1+1/2+1/3+1/4+...+1/100)-(1+1/2+1/3+..+1/50)
=1/51+1/52+1/53+..+1/100 (1)
A=1/51+1/52+1/53+..+1/100 (2)
(1),(2)=> A/B=1
NT
2
13 tháng 10 2018
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\left(đpcm\right)\)
13 tháng 10 2018
Ta có : \(VT=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
\(\Rightarrow\) \(ĐPCM\)
NT
1
20 tháng 6 2020
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!!!!
DQ
1
S
7 tháng 3 2019
\(C=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\Rightarrow C:D=1\)
