Đáp án:
`a, | x + 1 | + | x + 2 | + | x + 3 | + .... + | x + 2017 | = 2018x`
`⇒x+1+x+2+x+3+....+x+2017=2018x`
`⇒(x+x+x+...+x)+(1+2+3+...+2017)=2018x`
`⇒2017x+2035153=2018x`
`⇒x=2035153`
Vậy `x=2035153`
`b, | x^2 + 4x | + | x^2 - 16 | = 0`
`⇒ | x^2 + 4x | + | x^2 - 16 | = 0`
`⇒`\(\left[ \begin{array}{l}| x^2 + 4x | = 0\\| x^2 - 16 |=0\end{array} \right.\)
`⇒ x^2+4x=0`
`⇒x.(x+4)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x+4=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
`| x^2 - 16 |=0`
`⇒ x^2-16=0`
`x^2 - 16 = x^2 - 4^2 `
`⇒ (x - 4)(x + 4) = 0`
`⇒`\(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`⇔x=4`
