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|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
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
Bài 4:
\(\left|x+1\right|+\left|2x-3\right|=x-2\)
\(\Leftrightarrow x-2=\left|x+1\right|+\left|2x-3\right|\ge0\)
\(\Leftrightarrow x\ge2\)
\(\Leftrightarrow x+1>0\Leftrightarrow\left|x+1\right|=x+1\)
\(\Leftrightarrow2x-3>0\Leftrightarrow\left|2x-3\right|=2x-3\)
Lúc đó:
\(x+1+2x-3=x-2\)
\(\Leftrightarrow3x-2=x-2\Leftrightarrow x=0\)(Vô lý)
Bài 5:
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=5\)
Trường hợp 1: \(x\ge3\)
\(\left|x-1\right|=x-1\)
\(\left|x-2\right|=x-2\)
\(\left|x-3\right|=x-3\)
Lúc đó:
\(x-1+x-2+x-3=5\)
\(\Leftrightarrow3x-6=5\Leftrightarrow x=\frac{11}{3}\)(Thỏa mãn)
Trường hợp 2: \(2\le x\le3\)
\(\left|x-1\right|=x-1\)
\(\left|x-2\right|=x-2\)
\(\left|x-3\right|=3-x\)
Lúc đó:
\(x-1+x-2+3-x=5\)
\(\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)(Thỏa mãn)
Trường hợp 3:\(1\le x\le2\)
\(\left|x-1\right|x=x-1\)
\(\left|x-2\right|=2-x\)
\(\left|x-3\right|=3-x\)
Lúc đó:
\(x-1+2-x+3-x=5\)
\(\Leftrightarrow4-x=5\Leftrightarrow x=\left(-1\right)\)(Loại)
Trường hợp 4: \(x< 1\)
\(\left|x-1\right|=1-x\)
\(\left|x-2\right|=2-x\)
\(\left|x-3\right|=3-x\)
Lúc đó:
\(1-x+2-x+3-x=5\)
\(\Leftrightarrow6-3x=5\Leftrightarrow x=\frac{1}{3}\)(Thỏa mãn)
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