3(2y+1)=2(1+y)+3(3y+3) 解方程

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有三道化简题,大家帮忙啊1.(x-2y+1)^2-(x+2y-3分之2)(x-2y+3分之2)2.(4+a)(16-4a
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2.(4+a)(16-4a+a^2)-(2-a)^3
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1.(x-2y+1)^2-(x+2y-3分之2)(x-2y+3分之2)
=(x-2y+1)^2-[x+(2y-3分之2)][x-(2y-3分之2)]
=(x-2y+1)^2-[x^2-(2y-3分之2)^2]
=x^2-2x(2y-1)+(2y-1)^2-x^2+(2y-3分之2)^2
=(2y-1)^2-2x(2y-1)+(2y-3分之2)^2
=4y^2-4y+1-4xy+2x+4y^2-8y/3+4/9
=8y^2-20y/3-4xy+2x+13/9
2.(4+a)(16-4a+a^2)-(2-a)^3
=(4+a)[12+(2-a)^2]-(2-a)^3
=12(4+a)+(4+a)(2-a)^2-(2-a)^3
=12(4+a)+(4+a-2+a)(2-a)^2
=12(4+a)+2(1+a)(2-a)^2
3.(x-y)[(x+y)^2-xy](x^3+y^3)-(x+y)^3(x-y)^3
=(x-y)(x+y)^2(x^3+y^3)-(x-y)xy(x^3+y^3)-(x+y)^3(x-y)^3
=(x-y)(x+y)^2[(x^3+y^3)-(x+y)(x-y)^2]-(x-y)xy(x^3+y^3)
=(x-y)(x+y)^2[x^3+y^3-(x^2-y^2)(x-y)]-(x-y)xy(x^3+y^3)
=(x-y)(x+y)^2[x^3+y^3-(x^3-x^2y-xy^2+y^3)]-(x-y)xy(x^3+y^3)
=(x-y)(x+y)^2(x^2y+xy^2)-(x-y)xy(x^3+y^3)
=(x-y)xy(x+y)^3-(x-y)xy(x^3+y^3)
=(x-y)xy(x^3+3x^2y+3xy^2+y^3-x^3-y^3)
=(x-y)xy(3x^2y+3xy^2)
=3x^2y^2(x+y)(x-y)
用因式分解法.,(2y+1)(y-2)=0,
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倾斜角为θ,θ≠π/2时,
斜率k=tanθ
本例:
斜率k=(2y+1+3)/(4-2)=tan3π/4=-1
∴2y+4=-2
y=-3