Giải:
a) x2+5x=6
⇔x2+5x−6=0
⇔x2+6x−x−6=0
⇔(x2+6x)−(x+6)=0
⇔x(x+6)−(x+6)=0
⇔(x+6)(x−1)=0
⇔[x+6=0x−1=0⇔[x=−6x=1
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b) x2−2015x+2014=0
⇔x2−2014x−x+2014=0
⇔(x2−2014x)−(x−2014)=0
⇔x(x−2014)−(x−2014)=0
⇔(x−2014)(x−1)=0
⇔[x−2014=0x−1=0⇔[x=2014x=1
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