(1+1/2+1/3+........+1/205)*x+2015=2016/1+2017/2+2018/3+......+4030/2015
tìm x
Những câu hỏi liên quan
cho A= 1/2015+2/2016+3/2017+...+2016/4030-2016
B=1/2015+1/2016+1/2017+...+1/4030. chứng minh A/B là số nguyên.
Giup mình với nha
Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016
A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016
A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016
A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016
A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016
A=2016 - 2014.(1/2015+1/2016+....+1/4030) -2016
A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)
A=-2014.(1/2015+1/2016+....+1/4030)
mà B = 1/2015+1/2016+....+1/4030
nên A : B = -2014
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các bn hãy ủng hộ mk nhé !!! Thanks everyone!!!
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Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4029}{2014}+\frac{4030}{2015}\)
26 tháng 4 2015 lúc 13:54
Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\).
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\)
=> x = 2015
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5 tháng 5 2015 lúc 21:31
Tìm x biết \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x+2015=\frac{2016}{1}+\frac{2017}{2}+\frac{2018}{3}+...+\frac{4030}{2015}\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)\(\Rightarrow x=2015\)
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tìm x biết
\(\left(1+\frac{1}{2}+\frac{1}{3}+...........+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+............+\frac{4030}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\Rightarrow x=2015\)
Bạn có thể tham khảo nhé!^-^
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tìm x biết \(\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2015}\right)\times x+2015=\frac{2016}{1}+\frac{2017}{2}+.....+\frac{4029}{2014}+\frac{4030}{2015}\)
Tìm x biết \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\). x + 2015 = \(\frac{2016}{1}+\frac{2017}{2}+...\frac{4029}{2014}+\frac{4030}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(1-\frac{2016}{1}\right)+\left(1-\frac{2017}{2}\right)+...+\left(1-\frac{4030}{2015}\right)\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\Rightarrow x=2015\)
Không hiểu thì hỏi mình nhé! Thiên dâng bữa nay chăm chỉ đột xuất ta???
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Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Cho A= \(\frac{1}{2015}+\frac{2}{2016}+\frac{3}{2017}+...................+\frac{2016}{4030}-2016\) và B= \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+...................+\frac{1}{4030}\) . Chứng minh rằng \(\frac{A}{B}\) là một số nguyên