Tìm x biết :
| x + 1 | + | x + 2 | + | x + 3 | + .......+ | x + 2014 | =2015x
tìm x, biết :
|x+1| + |x+2| + |x+3| +...+ |x+2014| = 2015x
=> x là số nguyên dượng
=> x+1+x+2+x+3+x+4+...+x+2014=|x+1|+|x+2|+|x+3|+|x+4|+...+|x+2014|=2015
=> 2014x + (1+2+3+4+...+2014)=2015x
=> 1+2+3+4+...+2014=x
tìm x biết:
|x+1|+|x+2|+|x+3|+|x+4|+....+|x+2014| = 2015x
tìm x biết: |x+1|+|x+2|+|x+3|+..........+|x+2014|=2015x
|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
Tìm x biết
[x+1]+[x+2]+[x+3]+.....+[x+2014]=2015x
Mai mình kiểm tra.Mong bạn giúp
Tìm x biết :| x+1|+|x+2|+|x+3|+|...+|x+2014|=2015x
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 \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2014\right|=2015x\Rightarrow2015x\ge0\Rightarrow x\ge0\)\(\Rightarrow x+1+x+2+....+x+2014=2015x\)
\(\Rightarrow2014x+\frac{2014.2015}{2}=2015x\)
\(\Rightarrow2014x+2029105=2015x\)
\(\Rightarrow2015x-2014x=2029105\)
\(\Rightarrow x=2029105\)
tìm x biết: (x-2015)2015x - (x-2015)(2015x+2015) +32015 - 3 = 2(31+32+33+34+......+32014)
tinh nhanh bieu thuc sau
a) x^6-20x^5-20x^4-20x^4-20x^3-20x^2-20x+1992 voi x= 21
b)x^7-26x^6+27x^5-47x^4-77x^3+50x^2+x+1989 tai x = 25
c)4029-2015x+2015x^2-....+2015x^2014-x^2015 tai x= 2014
d)A= (x-1/x-2015+x-2/x-2014+x-3/x-2013+....+x-2014/x-2+x-2015/x-1):(1/2 +1/3+1/4+....+1/2016) tai x=2016 dau / o day la phan do nha!!
xin nho cac ban giai giup cho minh! cam on nhieu!!!!!!!! minh dang can ket qua gap
Tìm x biết rằng (2015x−2014)3=8(x−1)3+(2013x−2012)3.
Tìm x biết rằng (2015x−2014)3=8(x−1)3+(2013x−2012)3.
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....