Tìm x biết: x+1/2014 + x+2/2013 + x+3/2012 = x+10/2005 + x+11/2004 + x+12/2003
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
28 tháng 9 2020
Đề đúng phải là:
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)
Cộng mỗi phân thức thêm 1, quy đồng rồi chuyển sang 1 vế ta được:
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+2015}{2003}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà BT tích sau luôn nhỏ hơn 0
=> x+2015=0 => x = -2015
28 tháng 9 2020
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)( như này đúng không ? :)) )
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+10}{2005}+1\right)+\left(\frac{x+11}{2004}+1\right)+\left(\frac{x+12}{2003}+1\right)\)
<=> \(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}=\frac{x+10+2005}{2005}+\frac{x+11+2004}{2004}+\frac{x+12+2003}{2003}\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=\frac{x+2015}{2005}+\frac{x+2015}{2004}+\frac{x+12}{2003}\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+12}{2003}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)
=> x + 2015 = 0
=> x = -2015
5 tháng 12 2015
a) (x-5)x+2015 - (x-5)x+2014 =0
(x-5)x+2014(x-5 -1) =0
+ x -5 =0 => x =5
+ x -6 =0 => x =6
Vậy x = 5 hoặc x =6
QN
0
TN
25 tháng 8 2017
a) \(\frac{x-1}{2015}+\frac{x-2}{2014}=\frac{x-3}{2013}+\frac{x-4}{2012}\)
\(\Rightarrow\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)=\left(\frac{x-3}{2013}-1\right)+\left(\frac{x-4}{2012}-1\right)\)
\(\Rightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}=\frac{x-2016}{2013}+\frac{x-2016}{2012}\)
\(\Rightarrow\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)
\(\Rightarrow\left(x-2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\Rightarrow x-2016=0\)
\(\Rightarrow x=2016\)
b) \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
\(\Rightarrow\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Rightarrow\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
vì \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
c) \(|5x-3|\ge7\)
\(\Rightarrow5x-3\ge7\) hoặc - (5x-3) \(\ge7\)
\(\Rightarrow5x-3\ge7\) hoặc \(-5x+3\ge7\)
\(\Rightarrow5x\ge10\) hoặc \(-5x\ge4\)
\(\Rightarrow x\ge2\) hoặc \(x\le\frac{4}{-5}\)
k nhé!!! Kp luôn nha!
DT
0