2012/51+2012/52+2012/53+...+2012/100
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Những câu hỏi liên quan
1. 1+1/2+1/3+...+1/2013
2. 2014/1+2015/2+...+4025/2012+4026/2013
3. 2012/51+2012/52+2012/53+...+2012/100
Mong các bạn giúp mình, cảm ơn!
Tìm x biết (1/1x2+1/3x4+…+1/99x100)xX=2012/51+2012/52+…+2012/99+2012/100.
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
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Giải phương trình: \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)2013x=\dfrac{2012}{51}+\dfrac{2012}{52}+\dfrac{2012}{99}+\dfrac{2012}{100}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(2013x.\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
=> \(2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\Rightarrow2013x=2012\Rightarrow x=\frac{2012}{2013}\)
Vậy \(x=\frac{2012}{2013}\)
p/s: --trình bày sai sót mong bỏ qua
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Giải phương trình: \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)2013x=\dfrac{2012}{51}+\dfrac{2012}{52}+\dfrac{2012}{99}+\dfrac{2012}{100}\)
tim x biet:
: \(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
Xét vế trái biểu thức, ta có:
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot x\)
\(=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\right]\cdot x\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x\)
Xét vế phải biểu thức, ta có:
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
Từ đầu bài và 2 kết luận trên, ta suy ra:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)
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Xem thêm câu trả lời
Tìm x, biết: (1/1.2+1/3.4+...+1/99.100).x=2012/51+2012/52+...+2012/99+2012/100
giúp mình nhaaaaaaaaaaaaaaaaaaa ^^ (cho bạn luông 1 0 tick í í í ^.*)
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
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so sanh :
a=2010 /2011 +2011/2012 +2012/2013
b =2010+2011+2012 /2011 +20120 +2012
so sanh 1/51+1/52 +1/53 +.......+1/60 voi:
a .1/5
b.1/6
tra loi nhanh ho minh nha !
a < b k mình nha xong mình k lại cho
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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
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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
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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
Trần Thị Huệ
tìm x
(\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)). x =\(\frac{2012}{51}\)+\(\frac{2012}{52}\)+....+\(\frac{2012}{99}\)+\(\frac{2012}{100}\)