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
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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.2 B.-2 C.1/2 D.-1/2
3.要使式子4a^2-12a成为一个完全平方式的结果,应加上()
A.3 B.9 C.2.25 D.1.5
4.若x-y=3,x-z=1,则(2x-y-z)^2+(z-x)^2的值是()
A.16 B.17 C.4 D.2
5.(x+y-z)(x-y+z)=【 +( )】【 -( )】=( )^2-( )^2=
6.(3x-5y)(10y-6x)= ( )^2=
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.2 B.-2 C.1/2 D.-1/2
3.要使式子4a^2-12a成为一个完全平方式的结果,应加上()
A.3 B.9 C.2.25 D.1.5
4.若x-y=3,x-z=1,则(2x-y-z)^2+(z-x)^2的值是()
A.16 B.17 C.4 D.2
5.(x+y-z)(x-y+z)=【 +( )】【 -( )】=( )^2-( )^2=
6.(3x-5y)(10y-6x)= ( )^2=
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
→(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
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