(1)计算:(4xy-x2-y2)-(x2-y2+6xy)(2)计算:x(x-1)+(2x+5)(2x-5)(3)已知2x=y+15,求[(x2+y2)-(x-
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(1)计算:(4xy-x2-y2)-(x2-y2+6xy) (2)计算:x(x-1)+(2x+5)(2x-5) (3)已知2x=y+15,求[(x2+y2)-(x-y)2+2y(x-y)]÷2y的值. |
答案
(1)(4xy-x2-y2)-(x2-y2+6xy) =4xy-x2-y2-x2+y2-6xy =(-1-1)x2+(-1+1)y2+(4-6)xy =-2x2-2xy;
(2)x(x-1)+(2x+5)(2x-5) =x2-x+4x2-25 =5x2-x-25;
(3)[(x2+y2)-(x-y)2+2y(x-y)]÷2y =[x2+y2-(x2-2xy+y2)+2xy-2y2]÷2y =(x2+y2-x2+2xy-y2+2xy-2y2)÷2y =(-2y2+4xy)÷2y =-y+2x…(2分) 由于2x=y+15,则-y+2x=15,代入原式=15. |
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