Tìm x biết (1/1x2+1/3x4+…+1/99x100)xX=2012/51+2012/52+…+2012/99+2012/100.
Những câu hỏi liên quan
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, 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
Đúng 1
Bình luận (0)
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:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
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
Đúng 0
Bình luận (0)
Giải phương trình: \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)2013x=\dfrac{2012}{51}+\dfrac{2012}{52}+\dfrac{2012}{99}+\dfrac{2012}{100}\)
tìm x
(\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)). x =\(\frac{2012}{51}\)+\(\frac{2012}{52}\)+....+\(\frac{2012}{99}\)+\(\frac{2012}{100}\)
2012/51+2012/52+2012/53+...+2012/100
a) dfrac{2}{1^2}.dfrac{6}{2^2}.dfrac{12}{3^2}.dfrac{20}{4^2}.dfrac{30}{5^2}.....dfrac{110}{10^2}.x-20
b) left(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{2013}right).x+2013dfrac{2014}{1}+dfrac{2015}{2}+...+dfrac{4025}{2012}+dfrac{4026}{2013}
c) left(dfrac{1}{1.2}+dfrac{1}{3.4}+...+dfrac{1}{99.100}right).xdfrac{2012}{51}+dfrac{2012}{52}+...+dfrac{2012}{99}+dfrac{2012}{100}
Đọc tiếp
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
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!