`e)`
`3(x-1/2)-5(x+3/5)=-x+1/5`
`=>3x-3/2-5x-3+x-1/5=0`
`=>(3x-5x+x)+(3/2-3-1/5)=0`
`=>-x-47/10=0`
`=>-x=47/10`
`=>x=-47/10`
vậy `x=-47/10`
`f)`
`(2x-1)(x+2/3)=0`
`<=>` \(\left[ \begin{array}{l}2x-1=0\\x+\dfrac{2}{3}=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}2x=1\\x=-\dfrac{2}{3}\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{array} \right.\)
vậy `x∈{1/2;-2/3}`
`g)`
`(x+4)/2008+(x+3)/2009=(x+2)/2010+(x+1)/2011`
`=>2+(x+4)/2008+(x+3)/2009=2+(x+2)/2010+(x+1)/2011`
`=>1+(x+4)/2008+1+(x+3)/2009=1+(x+2)/2010+1+(x+1)/2011`
`=>(x+4+2008)/2008 +(x+3+2009)/2009=(x+2+2010)/2010+(x+1+2011)/2011`
`=> (x+2012)/2008+(x+2012)/2009=(x+2012)/2010+(x+2012)/2011`
`=> (x+2012)/2008+(x+2012)/2009-(x+2012)/2010-(x+2012)/2011=0`
`=>(x+2012)(1/2008+1/2009-1/2010-1/2011)=0`
vì `1/2008+1/2009-1/2010-1/2011 ne 0`
do đó `x+2012=0`
`=>x=-2012`
vậy `x=-2012`
