`a)|2x+1|=5`
`⇔`\(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
Vậy $x={-3;2}$
`b)2/(2.3)+2/(3.4)+2/(4.5)+....+2/(x.(x+1))=2008/2009`
`⇔ 2 . ( 1/(2.3) + 1/(3.4) + 1/(4.5) + ... + 1/(x(x+1)))=2008/2009`
`⇔ 2 . ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/(x+1) )=2008/2009`
`⇔ 2 . ( 1/2 - 1/(x+1) )=2008/2009`
`⇔ 1/2 - 1/(x+1) =2008/4018`
`⇔1/(x+1)=1/4018`
`⇔x+1=4018`
`⇔x=4017`
Vậy `x=4017`
