1.x^4-2=(x-y)(x+y)(x^2+y^2),当x=9.y=9时.x-y=0.x+y=18.x^2+y^2=162.则连起来为018162,对于多项式:4x^3-xy^2取x=10.y=10用上述方法可得到密码有()()()三个..2.说明任意两个连续奇数的平方差能被8整
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1.x^4-2=(x-y)(x+y)(x^2+y^2),当x=9.y=9时.x-y=0.x+y=18.x^2+y^2=162.则连起来为018162,对于多项式:4x^3-xy^2取x=10.y=10
1.x^4-2=(x-y)(x+y)(x^2+y^2),当x=9.y=9时.x-y=0.x+y=18.x^2+y^2=162.则连起来为018162,对于多项式:4x^3-xy^2取x=10.y=10用上述方法可得到密码有()()()三个..2.说明任意两个连续奇数的平方差能被8整
1.x^4-2=(x-y)(x+y)(x^2+y^2),当x=9.y=9时.x-y=0.x+y=18.x^2+y^2=162.则连起来为018162,对于多项式:4x^3-xy^2取x=10.y=10用上述方法可得到密码有()()()三个..
2.说明任意两个连续奇数的平方差能被8整除
1.x^4-2=(x-y)(x+y)(x^2+y^2),当x=9.y=9时.x-y=0.x+y=18.x^2+y^2=162.则连起来为018162,对于多项式:4x^3-xy^2取x=10.y=10用上述方法可得到密码有()()()三个..2.说明任意两个连续奇数的平方差能被8整
1)因为4x^3-xy^2=x(4x^2-y^2)=x(2x+y)(2x-y),
当x=10.y=10时,2x+y=30,2x-y=10,
用上述方法可得到密码有:301010,103010,101030三个
2)设任意两个连续奇数为2k+1,2k+3,(k为正整数)
因为(2k+3)^2-(2k+1)^2
=(2k+3+2k+1)(2k+3-2k-1)
=(4k+4)*2
=8(k+1)
所以任意两个连续奇数的平方差能被8整除
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2分之x+y+3分之x-y;4(x+y)-(x-y)谢谢了,
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