a) (x-3)/2014 +(x-2)/2015=(x-1)/2016 -3x²/(x²-1)+3/(x²-1)
⇔ (x-3)/2014 -1+(x-2)/2015-1-(x-1)/2016 -1=-3x²/(x²-1)+3/(x²-1) +3
⇔(x-2017)/2014 +(x-2017)/2015-(x-2017)/2016 =(3-3x²)/(x^2+1)
⇔(x-2017)/2014 +(x-2017)/2015-(x-2017)/2016=-3(x^2-1)/(x^2-1)+3
⇔(x-2017)/2014 +(x-2017)/2015-(x-2017)/2016=-3+3=0
⇔(x-2017(1/2014+1/2015 -1/2016)=0
⇔x-2017=0⇔x=2017 Vậy ..........
b)ĐKXĐ :x khác 0,b khác 0,a khác 0 và x khác -a-b
1/x+1/a+1/b=1/(x+a+b)
⇔1/a+1/b=1/(x+a+b)-1/x
⇔(a+b)/ab=(x-x-a-b)/(x²+ax+bx)
⇔(a+b)/ab=-(a+b)/(x²+ax+bx)
⇔(a+b)(x²+ax+bx)=-(a+b)ab
⇔-ab=x²+ax+bx
⇔-ab-x²-ax-bx=0
⇔-(ab+ax)-(x²+bx)=0
⇔-a(b+x)-x(b+x)=0
⇔-(a+x)(x+b)=0
⇔x=-a hoặc x=-b Vậy ..........
