初一 数学 初一数学几道计算题 请详细解答,谢谢! (27 14:42:32)

1.先化简,再求值: (2X+Y)^2-(2X-Y)(X+Y)-2(X-2Y)(X+2Y),其中X=0.5,Y=-2
2.已知X^2+Y^2-2X+2Y=-2,求代数值X^2008+Y^2009的值
3.A-(A^2B-2A^3B^2)\AB
4.0.5^(-1)-4*(-2)^(-2)+0.5^0-2^(-1)

1.先化简,再求值: (2X+Y)^2-(2X-Y)(X+Y)-2(X-2Y)(X+2Y),其中X=0.5,Y=-2

原式=4x^2+4xy+y^2-(2x^2+2xy-xy-y^2)-2(x^2-4y^2)
=3xy+10y^2

=3*0.5*(-2)+10*(-2)^2
=-3+40
=37

2.已知X^2+Y^2-2X+2Y=-2,求代数值X^2008+Y^2009的值

(x^2-2x+1)+(y^2+2y+1)=0
(x-1)^2+(y+1)^2=0
x-1=0
y+1=0

x=1,y=-1
x^2008+y^2009=1-1=0

3.A-(A^2B-2A^3B^2)\AB
=A-AB(A-2A^2B)/AB
=A-(A-2A^2B)
=2A^2B

4.0.5^(-1)-4*(-2)^(-2)+0.5^0-2^(-1)
=2-4*1/4+1-1/2
=2-1+1/2
=3/2

1、
(2x+y)^2-(2x-y)(x+y)-2(x-2y)(x+2y)
原式=4x^2+4xy+y^2-2x^2+xy-2xy+y^2-2x^2+8y^2
=3xy+10y^2

解：把x=0.5,y=-2
原式=3xy+10y^2
=-3+40
=37

2、解：
x^2+y^2-2x+2y=-2
(x^2-2x+1)+(y^2+2y+1)=0
(x-1)^2+(y+1)^2=0
所以 x=1 y=-1
所以 x^2008+y^2009=1-1=0

3、
A-[(A^2)B-2(A^3)(B^2)]/AB
=A-AB[A-2(A^2)B]/AB
=A-A+2(A^2)B
=2(A^2)B