Question

先化简,在求值:(3x 2y)(3x-2y)-(x-y)^2-2xy.其中x=1/2,y=2

Answer 1

先化简,在求值:(3x+2y)(3x-2y)-(x-y)^2-2xy.其中x=1/2,y=2
(3x+2y)(3x-2y)-(x-y)^2-2xy
解，得：
==3x^2-4y^2-(x^2-2xy+y^2)-2xy
==3x^2-4y^2-x^2+2xy-y^2-2xy
==2x^2-5y^2
把x=1/2,y=2带入2x^2-5y^2，得：
2*1^2-5*2^2
==2-20
==-18

Answer 2

原式=（3x)^2-(2y)^2-[(x-y)^2+2xy]
=9x^2- 4y^2-[x^2+y^2-2xy+2xy]
=9x^2-4y^2-x^2-y^2
=8x^2-5y^2
代入x=1/2,y=2
原式=8* 1/4 -5*2*2
=2-20
=-18