\(\frac{x-3}{2013}+\frac{x+1}{2014}=\frac{x-4025}{2}+\frac{x-4023}{3}\)tìm x
Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right).x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right).x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
(\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\) ) \(.x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
Ta có
\(\frac{2014}{1}+\frac{2015}{2}+...+\frac{4026}{2013}=1+1+...+1+\left[\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4026}{2013}-1\right)\right]\)
\(=2013+\left(\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2013}\right)=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\) (1)
Ta kết hợp (1) và đề
=>\(\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)x+2013=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\)
=> x=2013
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4025}{2012}-1\right)+\left(\frac{4026}{2013}-1\right)\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=2013\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
\(\Rightarrow x=\frac{2013\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}=2013\)
Vậy x = 2013 thoả mãn đề bài.
\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
tìm x
(1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+...+\(\frac{1}{2013}\)).x +2013=\(\frac{2014}{1}\)+\(\frac{2015}{2}\)+...+\(\frac{4025}{2012}\)+\(\frac{4026}{2013}\)
Tìm x biết
\(\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2014}+\frac{1}{2015}\right).x=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
Tìm x biết :
a ) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\left(\frac{2}{3}x+\frac{3}{4}\right)< 0\)
b) \(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+2}{2014}\)
tìm x : \(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)
=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0
=> x + 2017 = 0
=> x = -2017
\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)
\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)
\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Dễ thấy cái ngoặc to < 0
=> x=-2017
tìm x biết :(\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}.x=\frac{2013}{1}+\frac{2012}{2}+....+\frac{2}{2012}+\frac{1}{2013}\)