1.[(2/3)x^6y^3]/(-2xy)^2-(-x/3)^2*x^2y=[(2/3)x^6y^3]/(4x^2y^2)-(x^2/9)*x^2y=(1/6)x^4y-(1/9)x^4y=(1/18)x^4y2.(3x+1)(2y-4)=3x*2y-3x*4+1*2y-4=6xy-12x+2y-43.(2x-1)^2(2x+1)^2=[(2x-1)(2x+1)]^2=[4x^2-1]^2=16 展开
1.[(2/3)x^6y^3]/(-2xy)^2-(-x/3)^2*x^2y=[(2/3)x^6y^3]/(4x^2y^2)-(x^2/9)*x^2y=(1/6)x^4y-(1/9)x^4y=(1/18)x^4y2.(3x+1)(2y-4)=3x*2y-3x*4+1*2y-4=6xy-12x+2y-43.(2x-1)^2(2x+1)^2=[(2x-1)(2x+1)]^2=[4x^2-1]^2=16x^4-8x^2+14.(a^2-b)(a^4+a^2b+b^2)-a^2(a^4-b)=a^2*a^4+a^2*a^2b+a^2*b^2-b*a^4-b*a^2b-b*b^2-a^2*a^4+a^2*b=a^6+a^4b+a^2b^2-a^4b-a^2b^2-b^3-a^6+a^2b=a^2b-b^3分解因式 1.4y^2(1-m)-2y(m-1)=(2y)(2y)(1-m)+2y(1-m)=2y(1-m)(2y+1) 2.a^4-18a^2+81=(a^2)^2-2*9*a^2+9^2=(a^2-9)^2=[(a+3)(a-3)]^23.(x^2+2x)^2-2(x^2+2x)-3=(x^2+2x+1)(x^2+2x-3)=(x+1)^2(x-1)(x+3)4.（x+1)(x+5)(x+3)(x+7）-9 =[(x+1)(x+7)][(x+3)(x+5)]-9=(x^2+8x+7)(x^2+8x+15)-9设x^2+8x+7=y,则上式变为y(y+8)-9=y^2+8y-9=(y+9)(y-1)即原式=(x^2+8x+7+9)(x^2+8x+7-1)=(x^2+8x+16)(x^2+8x+6)=(x+4)^2(x^2+8x+6)简答题 已知：2x-4y=O,求(3x^2-y^2)/(x^2+2xy)2x-4y=0,所以x=2y,代入(3x^2-y^2)/(x^2+2xy)中得[3(2y)^2-y^2]/[(2y)^2+2*2y*y]=(12y^2-y^2)/(4y^2+4y^2)=11y^2/(8y^2)=11/8 收起
