Question

急求数学答案1.a^2+b^2-4a+6b+13=0,求b^a的值2.a^2+b^2=3,a-b=2,则ab的值是（）A

Answer 1

⑴ a^2+b^2-4a+6b+13=0

→(a-2)^2+(b+3)^2=0

即(a-2)^2=0,(b+3)^2=0

∴a=2,b=-3

∴a^b=2^(-3)=1/8

⑵∵a-b=2

∴(a-b)^2=4

既a^2+b^2-2ab=4

把a^2+b^2=3代入上式得:

3+2ab=4

ab=1/2

既C

⑶原式=4(a^2-3a)

=4(a^2-3a+(3/2)^2-(3/2)^2)

=4(a-3/2)^2-9

既应加上9所以选B

⑷解x-y=3得x=3+y①

解x-z=1得x=1+z②

把①,②代入原式得

(x+x-y-z)^+(z-x)^2

=((3+y)+(1+z)-y-z)^2+(z-(1+z))^2

=(3+y+1+z-y-z)^2+(z-1-z)^2

=(3+1)^2+(-1)^2

=4^2+1

=16+1

=17

即选B

⑸ (x+y-z)(x-y+z)

=x^2-xy+xz+xy-y^2+yz-xz+yz-z^2

=x^2-y^2+2yz-z^2

=x^2-(y^2-2yz+z^2)

=x^2-(y-z)^2

⑹ (3x-5y)(10y-6x)

=(3x-5y)(-2(3x-5y))

=-2(3x-5y)(3x-5y)

=-2(3x-5y)^2