tìm x biết: |x+1|+|x+2|+|x+3|+..........+|x+2014|=2015x
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
ND
1
PV
0
DN
4
16 tháng 2 2017
Ta thấy :
\(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
............
|x + 2014| \(\ge0\)
Cộng vế với vế ta được :
\(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|\ge0\)
Mà \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|=2015x\Rightarrow2015x\ge0\Rightarrow x\ge0\)\(\Rightarrow x+1+x+2+....+x+2014=2015x\)
\(\Rightarrow2014x+\frac{2014.2015}{2}=2015x\)
\(\Rightarrow2014x+2029105=2015x\)
\(\Rightarrow2015x-2014x=2029105\)
\(\Rightarrow x=2029105\)
DT
1
18 tháng 3 2017
Ta có:
\(\left|x+1\right|\ge0,\left|x+2\right|\ge0,...,\left|x+2014\right|\ge0\)
\(\Rightarrow\)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2014\right|\ge0\)
\(\Rightarrow2015x\ge0\)
\(\Rightarrow x\ge0\)
Khi đó :\(\left|x+1\right|=x+1,\left|x+2\right|=x+2,...,\left|x+2014\right|=x+2014\)\(\Rightarrow x+1+x+2+...+x+2014=2015x\)
\(\Rightarrow2014x+1+2+...+2014=2015x\)
\(\Rightarrow1+2+..+2014=x\)
\(\Rightarrow x=\dfrac{\left(1+2014\right)2014}{2}=2029105\)
FZ
0
FZ
1
15 tháng 10 2015
PT <=> (2015x - 2014)3 = (2x - 2)3 + (2013x - 2012)3
<=> (2015x - 2014)3 = (2x - 2 + 2013x - 2012). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]
<=> (2015x - 2014)3 = (2015x - 2014). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]
<=> (2015x - 2014).[ (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]] = 0
<=> 2015.x - 2014 = 0 hoặc (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0
+) 2015x - 2014 = 0 => x = 2014/2015
+) (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0
<=> [(2x - 2) + (2013x - 2012)]2 - (2x - 2)2 + (2x - 2).(2013x - 2012) - (2013x - 2012)2 = 0
<=> 3. (2x - 2).(2013x - 2012) = 0
<=> 2x - 2 = 0 hoặc 2013x - 2012 = 0
<=> x = 1 hoặc x = 2012/2013
Vậy....
|x+1| + |x+2| + |x+3| + .......... + |x+2014| = 2015x
Ta có :
|x+1| \(\ge\)0
|x+2| \(\ge\)0
|x+3| \(\ge\)0
..........
|x+2014| \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| \(\ge\)0
=> 2015x \(\ge\)0
Mà 2015 \(\ge\)0
=> x \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014|
= x + 1 + x + 2 + x + 3 +.................... + x + 2014 = 2015x
=> 2014x + (1 + 2 + 3 +............ + 2014) = 2015x
=> 1 + 2 + 3 + 4 + ........................ + 2014 = x
=> x = 2029105