tìm số nguyên a biết
( 1/1x2+1/3x4+.....+1/99x100) x a =2012/51+2012/52+....+2012/100
giải nhanh lên nhé các bạn,chi tiết đó nha
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ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
bài toán được viết lại như sau:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\)
vậy x=2012
Ta có:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> x = 2012
a)
Ta có a > b vì b > 3 còn a < 3
b)
a. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 < 1/51 x 10 < 1/50 x 10 = 1/5
=> 1/51 + 1/52 +1/53 +...+1/60 < 1/5
b. Ta có : 1/51 + 1/52 + 1/53 +...+ 1/60 > 1/60 x 10 = 1/6
=> 1/51 + 1/52 +1/53 +...+ 1/60 > 1/6
a; 19,29,59
b. 889=887+3 (887 nguyen to)
c.2001.2002.2003.2004 co tan cung la 4
vay 2001.2002.2003.2004 +1 co tan cung la 5
vay (c) luon chia het cho 5= hop so
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
Có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=+....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
= \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=1\)
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