Đặt A=|x + 1| + |x + 2| + |x + 3| + |x + 4| + |x + 5| = 2006x
Vì vế trái luôn \(\ge\)0 với mọi x
=>Vế phải luôn \(\ge\)0
=> 2006x \(\ge\) 0
=>x\(\ge\)0
=> x + 1 > 0; x + 2 > 0; x + 3 > 0; x + 4 > 0; x + 5 > 0
=> |x + 1| = x + 1; |x + 2| = x + 2; |x + 3| = x + 3; |x + 4| = x + 4; |x + 5| = x + 5
Khi đó A trở thành:
x+1+x+2+x+3+x+4+x+5=2006x
Ta có: 5x+15=2006x
15=2006x-5x
15=2001x
x=15/2001=5/667
Vậy x=5/667
|x+1|+|x+2|+|x+3|+|x+4|+|x+5|=2006x (1)
Vì |x+1| > 0 ;|x+2| > 0;|x+3| > 0;|x+4| > 0;|x+5| > 0
=>|x+1|+|x+2|+|x+3|+|x+4|+|x+5| > 0
=>2006x > 0=>x > 0
Do đó |x+1|=x+1;|x+2|=x+2;|x+3|=x+3;|x+4|=x+4;|x+5|=x+5
=> (1) trở thành : x+1+x+2+x+3+x+4+x+5=2006x
=>(x+x+x+x+x)+(1+2+3+4+5)=2006x
=>5x+15=2006x
=>2006x-5x=15=>2001x=15=>x=15/2001=5/667
Vậy x=5/667