|x+1|+|x+2|+|x+3|+...+|x+100|=605x
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
|x + 1| + |x + 2| + |x + 3| + ...... + |x + 100| = 605x
Có : |x + 1| + |x + 2| + |x + 3| + ...... + |x + 100| \(\ge\)0
=> 605x \(\ge\) 0
=> x \(\ge\)0
<=> x + 1 + x + 2 + x + 3 + ...... + x + 100 = 605x
<=> 100x + 5050 = 605x
<=> 505x = 5050
<=> x = 10
|x + 1| + |x + 2| + |x + 3| + ...... + |x + 100| = 605x
Có : |x + 1| + |x + 2| + |x + 3| + ...... + |x + 100| \(\ge\)0
=> 605x \(\ge\) 0
=> x \(\ge\)0
<=> x + 1 + x + 2 + x + 3 + ...... + x + 100 = 605x
<=> 100x + 5050 = 605x
<=> 505x = 5050
<=> x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy :
| x + 1 | \(\ge\)0 ; | x + 2 | \(\ge\)0 ; ... ; | x + 100 | \(\ge\)0
\(\Rightarrow\)|x+1|+|x+2|+....+|x+100| \(\ge\)0 nên 605x \(\ge\)0
\(\Rightarrow\)( x + 1 ) + ( x + 2 ) + .... + ( x + 100 ) = 605x
( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 605x
100x + 5050 = 605x
505x = 5050
x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x+1\right|+\left|x+2\right|+...+\left|x+100\right|=605x\)(1)
Vì \(VT>0\forall x\)
\(\Rightarrow VP>0\Leftrightarrow605x>0\Leftrightarrow x>0\)
Khi đó \(\left(1\right)\Leftrightarrow x+1+x+2+...+x+100=605x\)
\(\Leftrightarrow100x+5050=605x\)
\(\Leftrightarrow505x=5050\)
\(\Leftrightarrow x=10\)( thỏa mãn )
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có 605x > 0 suy ra x>0
ta có x+1+x+2+x+...+100=605x
100x +5050=605x
505x=5050
x=10
Ta có \(|x+1|\ge0\)
\(|x+2|\ge0\)
\(|x+3|\ge0\)
\(.......\)
\(|x+100|\ge0\)
\(\Rightarrow|x+1|+|x+2|+|x+3|+...+|x+100|=605x\)
\(\Rightarrow x+1+x+2+x+3+...+x+100=605x\)
\(\Rightarrow\left(x+x+x+x+...+x\right)+\left(1+2+3+...+100\right)=605x\)
\(\Rightarrow100x+\left(100+1\right)\cdot100:2=605x\)
\(\Rightarrow100x+5050=605x\)
\(\Rightarrow605x-100x=5050\)
\(\Rightarrow x\left(605-100\right)=5050\)
\(\Rightarrow x\cdot505=5050\)
\(\Rightarrow x=5050:505\)
\(\Rightarrow x=10\)
Vậy x=10
Ta có:
|x+1|>=0 với mọi x
|x+2|>=0 với mọi x
.........................
|x+100|>=0 với mọi x
=> |x+1| + |x+2| + |x+3| + ......+ |x+100| >=0 với mọi x
Mà |x+1| + |x+2| + |x+3| + ......+ |x+100| = 605x
605x >=0 với mọi x
=> x>=0
Vậy |x+1|= x+1
|x+2|= x+2
.............
|x+100|= x+100
Vậy |x+1| + |x+2| + |x+3| + ......+ |x+100| = x+1+x+2 x+3 + ....+ x+100=605x
=100x+ 5050=605x
605x-100x=5050
505x=5050
=> x=10
Vậy X = 10