Cho A= 1/2 + 1/3 + 1/4 + …. + 1/100 + 1/101 B= 1/100.1 + 1/99.2 + 1/98.3 + …. + 1/52.49 + 1/51.50 09/08/2021 Bởi Nevaeh Cho A= 1/2 + 1/3 + 1/4 + …. + 1/100 + 1/101 B= 1/100.1 + 1/99.2 + 1/98.3 + …. + 1/52.49 + 1/51.50
Ta có : A = 1212 + 1313 + 1414 + ….. + 11001100 + 11011101 = 11001100 + 11011101 + ( 1212 + 199199) + ( 1313 + 198198 )+ …. +( 152152 + 149149) + ( 151151 + 150150) = 11001100 + 11011101 + 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50 => 101B = 101100101100 + 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50 B = 1100.11100.1 + 199.2199.2 + 198.3198.3+ …. + 152.49152.49 + 151.50151.50 => 101B =1 + 11001100+ 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50 = (11011101 – 1) + ( 11001100 – 11001100) + [(10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50) – (10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50)] = −100101 Mik k vt đc dạng phân số! #lehaanh Bình luận
Đáp án: Ta có : A = $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$ + ….. + $\dfrac{1}{100}$ + $\dfrac{1}{101}$ = $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + ( $\dfrac{1}{2}$ + $\dfrac{1}{99}$) + ( $\dfrac{1}{3}$ + $\dfrac{1}{98}$ )+ …. +( $\dfrac{1}{52}$ + $\dfrac{1}{49}$) + ( $\dfrac{1}{51}$ + $\dfrac{1}{50}$) = $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$ => 101B = $\dfrac{101}{100}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$ B = $\dfrac{1}{100.1}$ + $\dfrac{1}{99.2}$ + $\dfrac{1}{98.3}$+ …. + $\dfrac{1}{52.49}$ + $\dfrac{1}{51.50}$ => 101B =1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$ Ta có : A – 101B = ($\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ( 1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) = ($\dfrac{1}{101}$ – 1) + ( $\dfrac{1}{100}$ – $\dfrac{1}{100}$) + [($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$)] = $\dfrac{-100}{101}$ Giải thích các bước giải: Bình luận
Ta có :
A = 1212 + 1313 + 1414 + ….. + 11001100 + 11011101
= 11001100 + 11011101 + ( 1212 + 199199) + ( 1313 + 198198 )+ …. +( 152152 + 149149) + ( 151151 + 150150)
= 11001100 + 11011101 + 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50
=> 101B = 101100101100 + 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50
B = 1100.11100.1 + 199.2199.2 + 198.3198.3+ …. + 152.49152.49 + 151.50151.50
=> 101B =1 + 11001100+ 10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50
= (11011101 – 1) + ( 11001100 – 11001100) + [(10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50) – (10199.210199.2 + 10198.310198.3+ …. + 10152.4910152.49 + 10151.5010151.50)]
= −100101
Mik k vt đc dạng phân số!
#lehaanh
Đáp án:
Ta có :
A = $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$ + ….. + $\dfrac{1}{100}$ + $\dfrac{1}{101}$
= $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + ( $\dfrac{1}{2}$ + $\dfrac{1}{99}$) + ( $\dfrac{1}{3}$ + $\dfrac{1}{98}$ )+ …. +( $\dfrac{1}{52}$ + $\dfrac{1}{49}$) + ( $\dfrac{1}{51}$ + $\dfrac{1}{50}$)
= $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
=> 101B = $\dfrac{101}{100}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
B = $\dfrac{1}{100.1}$ + $\dfrac{1}{99.2}$ + $\dfrac{1}{98.3}$+ …. + $\dfrac{1}{52.49}$ + $\dfrac{1}{51.50}$
=> 101B =1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
Ta có :
A – 101B = ($\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ( 1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$)
= ($\dfrac{1}{101}$ – 1) + ( $\dfrac{1}{100}$ – $\dfrac{1}{100}$) + [($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$)]
= $\dfrac{-100}{101}$
Giải thích các bước giải: