∵tana和tanb是方程x^2-5x+6=0的两个实数根
∴tana+tanb=5,tanatanb=6
∴tan(a+b)=(tana+tanb)/(1-tanatanb)=5/(1-6)=-1
∴2[sin(a+b)]^2-3sin(a+b)cos(a+b)+[cos(a+b)]^2
={2[sin(a+b)]^2-3sin(a+b)cos(a+b)+[cos(a+b)]^2}/{[sin(a+b)]^2+[cos(a+b)]^2}
={2[tan(a+b)]^2-3tan(a+b)+1}/{[tan(a+b)]^2+1}
=(2+3+1)/(1+1)
=3.