初一乘法公式1.求证:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)=3(x+y-2z)(y+z-2x)(

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初一乘法公式
1.求证:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
2.在括号里填上适当的数或式,使等式成立
(1)x^2-1/2x+( )=(x-1/4)^2 (2)(a+b)^2+( )=(a-b)^2
(3)(1/2x-4)[1/4x^2+( )+16]=( )^3-( )^3
3.解方程.(5x+7)(5x-7)+(x+3)^2=10x^2-(4x+1)(-4x-1)+48
4.已知:(-4x+3y)(-3y-4x)减去多项式M的差为-36y^2+5xy,求M
5.已知：a+b+c=0,a^3+b^3+c^3=0求证a^19+b^19+c^19+0

1.求证:(x+y-2z)^3+(y+z-2x)^3+(y+z-2x)=3(x+y-2z)(y+z-2x)(z+x-2y)
令a=x+y-2z,b=y+z-2x,c=y+z-2x
则a+b+c=0
a^3+b^3+c^3-3abc
=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc
=(a+b)^3+c^3-3ab(a+b+c)
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
=0
所以a^3+b^3+c^3=3abc
即(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
2.在括号里填上适当的数或式,使等式成立
(1)x^2-1/2x+(1/16)=(x-1/4)^2 (2)(a+b)^2+(-4ab)=(a-b)^2
(3)(1/2x-4)[1/4x^2+(2x)+16]=(1/2x)^3-(4)^3
3.解方程.(5x+7)(5x-7)+(x+3)^2=10x^2-(4x+1)(-4x-1)+48
25x^2-49+x^2+6x+9=10x^2+16x^2+8x+1+48
-49+6x+9=8x+1+48
2x=-89
x=-89/2
4.已知:(-4x+3y)(-3y-4x)减去多项式M的差为-36y^2+5xy,求M
M=(-4x+3y)(-3y-4x)-(-36y^2+5xy)
=16x^2-9y^2+36y^2-5xy
=16x^2-5xy+27y^2
5.已知：a+b+c=0,a^3+b^3+c^3=0求证a^19+b^19+c^19=0
由第一题知道
a+b+c=0时,a^3+b^3+c^3=3abc
所以3abc=0
所以a,b,c至少有一个0
假设c=0
则a+b=0,
a=-b
a^19+b^19+c^19
=(-b)^19+b^19+0^19
=-b^19+b^19+0
=0

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