已知a+b=3,ab=1,则a²-ab+b²=.

已知a+b=3,ab=1,则a²-ab+b²=.
已知a+b=3,ab=1,则a²-ab+b²=.

已知a+b=3,ab=1,则a²-ab+b²=.
a²-ab+b²
=（a+b)^2-2ab-ab
=（a+b)^2-3ab
∵a+b=3,ab=1
∴原式=9-3=6

6

a²-ab+b²
=a²-ab+b²+3ab-3ab
=a²+2ab+b²-3ab
=(a+b)²-3ab
=3²-3*1
=9-3
=6

a²-ab+b²
=a²+2ab+b²-3ab
=(a+b)²-3ab
=3²-3×1
=9-3
=6

a+b=3,ab=1
aˆ2+bˆ2=(a+b)ˆ2-2ab=9-2=7
所以结果为7-1=6
可追问啊

(a+b)*(a+b)=a*a+b*b+2a*b=3*3=9
a²-ab+b²=(a+b)*(a+b)-3a*b=9-3=6

a²-ab+b²=a²+2ab+b²-3ab=(a+b)²-3ab=3²-3*1=9-3=6