已知x+y=4,xy=3,求:(1)x^2y-xy^2;(2)(x^2+1)(y^2+1)求解
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x+y=4,xy=3,所以(x-y)²=(x+y)²-4xy=16-4*3=4,所以x-y=±2
1、x^2y-xy^2=xy(x-y)=6(x>y时)或-6(x<y时)
2、(x^2+1)(y^2+1)=(xy)^2+(x^2+y^2)+1=9+1+(x+y)^2-2xy=10+16-6=20
1、x^2y-xy^2=xy(x-y)=6(x>y时)或-6(x<y时)
2、(x^2+1)(y^2+1)=(xy)^2+(x^2+y^2)+1=9+1+(x+y)^2-2xy=10+16-6=20
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x+y=4,xy=3,所以(x-y)²=(x+y)²-4xy=16-4*3=4,所以x-y=±2
1)x^2*y-x*y^2=xy(x-y)=±6;
2)(x^2+1)(y^2+1)=(xy)^2+x^2+y^2+1=(xy)^2+(x+y)^2-2xy+1=3^2+4^2-2*3+1=20
1)x^2*y-x*y^2=xy(x-y)=±6;
2)(x^2+1)(y^2+1)=(xy)^2+x^2+y^2+1=(xy)^2+(x+y)^2-2xy+1=3^2+4^2-2*3+1=20
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(x-y)^2=(x+y)^2-4xy=16-12=4
x-y=+-2
1)x^2y-xy^2
=xy(x-y)=+-6
(2)(x^2+1)(y^2+1)
=x^2t^2+x^2+y^2+1
=x^2y^2+(x+y)^2-2xy+1
=9+16-6+1
=20
x-y=+-2
1)x^2y-xy^2
=xy(x-y)=+-6
(2)(x^2+1)(y^2+1)
=x^2t^2+x^2+y^2+1
=x^2y^2+(x+y)^2-2xy+1
=9+16-6+1
=20
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x+y=4
xy=3
(1) (x-y)²=(x+y)²-4xy
=16-4*3
=4
x-y=±2
x²y-xy²=xy(x-y)
=±6
即 x²y-xy²=6 或 x²y-xy²=-6
(2) (x²+1)(y²+1)
=(xy)²+x²+y²+1
=(xy)²+(x+y)²-2xy+1
=9+16-2*3+1
=20
xy=3
(1) (x-y)²=(x+y)²-4xy
=16-4*3
=4
x-y=±2
x²y-xy²=xy(x-y)
=±6
即 x²y-xy²=6 或 x²y-xy²=-6
(2) (x²+1)(y²+1)
=(xy)²+x²+y²+1
=(xy)²+(x+y)²-2xy+1
=9+16-2*3+1
=20
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