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Ta có \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Lại có B = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{99.52}+\frac{1}{100.51}\)
=> 151B = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{99.52}+\frac{151}{100.51}\)
=> 151B = \(\frac{1}{51}+\frac{1}{100}+\frac{1}{52}+\frac{1}{99}+...+\frac{1}{99}+\frac{1}{52}+\frac{1}{100}+\frac{1}{51}\)
=> 151B = \(2\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> B = \(\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Khi đó \(\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{\frac{2}{151}}=\frac{151}{2}=75,5\)
xét B ta có:
B=1/1.2+1/3.4+1/5.6+...+1/99.100
B=1-1/2+1/3-1/4+1/5-1/6+...+1/99-100
B=(1+1/3+1/5+...+1/99)-(1/2+1/4+...+1/100)
B=(1+1/3+1/5+...+1/99)+(1/2+1/4+1/6+...+1/100)-2(1/2+1/4+1/6+...+1/100)
B=(1+1/2+1/3+...+1/99+1/100)-(1+1/2+1/3+1/4+...+1/50)
=>B=1/51+1/52+1/53+...+1/100
=>A/B=1/51+1/52+...+1/100:1/51+1/52+...+1/100=1 (đpcm)
Đó là cách nhanh nhất để giải nếu bn ko hỉu thì mik sẽ giải chi tiết cho
chúc bn học tốt ^-^
B = 1/1x2 + 1/3x4 + ... + 1/99x100
B = 1 - 1/2 + 1/3 - 1/4 + ... + 1/99 - 1/100
B = (1 + 1/2 + 1/3 + 1/4 + ... + 1/99 + 1/100) - (2.1/2 + 2.1/4 + 2.1/6 + ... + 2.1/100)
B = (1 + 1/2 + 1/3 + 1/4 + ... + 1/99 + 1/100) - (1 + 1/2 + 1/3 + ... + 1/50)
B = 1/51 + 1/52 + 1/53 + ... + 1/100
=> tỉ số a/b = 1
Ta có B =\(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{100.51}\)
=> 151B = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{100.51}=\frac{1}{51}+\frac{1}{100}+\frac{1}{52}+\frac{1}{99}+...+\frac{1}{51}+\frac{1}{100}\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> B = \(\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Khi đó \(\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{\frac{2}{151}}=\frac{151}{2}=75,5\)