Vì `|x+2|+|x+2/5|+|x+1/2|>=0`
`=>4x>=0`
`=>x>=0`
`=>|x+2|=x+2,|x+2/5|=x+2/5,|x+1/2|=x+1/2`
`=>x+2+x+2/5+x+1/2=4x`
`=>3x+5/2+2/5=4x`
`=>4x-3x=5/2+2/5=29/10`
`=>x=29/10(tmđk)`
Vậy `x=29/10`
Giải:
Vì \(\left|x+2\right|+\left|x+\dfrac{2}{5}\right|+\left|x+\dfrac{1}{2}\right|\ge0\) nên ta có:
\(\left|x+2\right|+\left|x+\dfrac{2}{5}\right|+\left|x+\dfrac{1}{2}\right|=4x\)
\(\left(x+2\right)+\left(x+\dfrac{2}{5}\right)+\left(x+\dfrac{1}{2}\right)=4x\)
\(x+2+x+\dfrac{2}{5}+x+\dfrac{1}{2}=4x\)
\(\left(x+x+x\right)+\left(2+\dfrac{2}{5}+\dfrac{1}{2}\right)=4x\)
\(3x+\dfrac{29}{10}=4x\)
\(3x-4x=-\dfrac{29}{10}\)
\(-1x=-\dfrac{29}{10}\)
\(x=-\dfrac{29}{10}:-1\)
\(x=\dfrac{29}{10}\)
