8x³-y³-18xy-27
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8x³-y³-18xy-27
x³-13x²-11x+23
x³+3x²-12x-36
4x³+x+15
27x³-9x²+28
abcd∈z,又m=(ab+cd)²-¼(a²+b²-c²-d²)为非零整数,问m的绝对值为合数吗
x³-13x²-11x+23
x³+3x²-12x-36
4x³+x+15
27x³-9x²+28
abcd∈z,又m=(ab+cd)²-¼(a²+b²-c²-d²)为非零整数,问m的绝对值为合数吗
1)此题需利用因式分解a³+b³+c³-3abc的公式
a³+b³+c³-3abc=(a+b)³+c³-3ab(a+b)-3abc
=(a+b+c)[(a+b)²-(a+b)c+c²-3ab]
=(a+b+c)(a²+b²+c²-ab-ac-bc)
∴8x³-y³-18xy-27
=(2x)³+(-y)³+(-3)³-3•(2x)(-y)(-3)
=(2x-y-3)(4x²+y²+9+2xy+6x-3y)
(2)x³-13x²-11x+23
=(x³-1)-(13x²+11x-24)
=(x-1)(x²+x+1)-(x-1)(13x+24)
=(x-1)(x²-12x-23)
(3)x³+3x²-12x-36
=x²(x+3)-12(x+3)
=(x+3)(x²-12)
先传这几题 其它几题我再研究
a³+b³+c³-3abc=(a+b)³+c³-3ab(a+b)-3abc
=(a+b+c)[(a+b)²-(a+b)c+c²-3ab]
=(a+b+c)(a²+b²+c²-ab-ac-bc)
∴8x³-y³-18xy-27
=(2x)³+(-y)³+(-3)³-3•(2x)(-y)(-3)
=(2x-y-3)(4x²+y²+9+2xy+6x-3y)
(2)x³-13x²-11x+23
=(x³-1)-(13x²+11x-24)
=(x-1)(x²+x+1)-(x-1)(13x+24)
=(x-1)(x²-12x-23)
(3)x³+3x²-12x-36
=x²(x+3)-12(x+3)
=(x+3)(x²-12)
先传这几题 其它几题我再研究
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