求求算下:解同余式f(x)≡3x^14+4x^13+2x^11+x^9++x^6+x^3+12x^2+x≡0(mod5)
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/18 21:15:20
求求算下:解同余式f(x)≡3x^14+4x^13+2x^11+x^9++x^6+x^3+12x^2+x≡0(mod5).
![求求算下:解同余式f(x)≡3x^14+4x^13+2x^11+x^9++x^6+x^3+12x^2+x≡0(mod5)](/uploads/image/z/8773533-45-3.jpg?t=%E6%B1%82%E6%B1%82%E7%AE%97%E4%B8%8B%EF%BC%9A%E8%A7%A3%E5%90%8C%E4%BD%99%E5%BC%8Ff%28x%29%E2%89%A13x%5E14%2B4x%5E13%2B2x%5E11%2Bx%5E9%2B%2Bx%5E6%2Bx%5E3%2B12x%5E2%2Bx%E2%89%A10%28mod5%29)
首先,x==0 mod 5是一解;
当x0 mod 5 时,由欧拉函数定理或费马小定理,x^4==1 mod 5,此时
f(x)
==3x^2 + 4x + 2x^3 + x + x^2 + x^3 + 12x^2 + x
=(3+1+12)xx+(4+1+1)x+(2+1)xxx
==xx+x+3xxx
=x(3xx+x+1)
==0 mod 5
由于x0 mod 5,故 3xx+x+1==0 mod 5
故6xx+2x+2==0==xx+2x+2
故(x+1)^2==-1 mod 5
故x==1,2 mod 5
综上,x==0,1,2 mod 5
当x0 mod 5 时,由欧拉函数定理或费马小定理,x^4==1 mod 5,此时
f(x)
==3x^2 + 4x + 2x^3 + x + x^2 + x^3 + 12x^2 + x
=(3+1+12)xx+(4+1+1)x+(2+1)xxx
==xx+x+3xxx
=x(3xx+x+1)
==0 mod 5
由于x0 mod 5,故 3xx+x+1==0 mod 5
故6xx+2x+2==0==xx+2x+2
故(x+1)^2==-1 mod 5
故x==1,2 mod 5
综上,x==0,1,2 mod 5
求求算下:解同余式f(x)≡3x^14+4x^13+2x^11+x^9++x^6+x^3+12x^2+x≡0(mod5)
1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x+13x+14x+15x=550
设函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10),
|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10|
x+2x+3x+4x+5x+6x+7x+8x+9x=9x-8x-7x-6x-5x-4x-3x-2x-x.x等于多少?
解同余式组x≡-2(mod12)x≡6(mod 10) x≡1(mod 15)
约分 (X的平方+4x+3)(2x-x)^/(X^+X)(X^+X-6),
解方程:(x+3)/(x+2)-(x+5)/(x+4)=(x+7)/(x+6)-(x+9)/(x+8)
已知 X=4- 根号3 求 (X*X*X*X-6X*X*X-2X*X+18X+23) / (X*X-8X+15)的值
解方程4x方+x+2x根号下(3x方+x) =9
f(x)=x(x-1)(x-2)(x-3)(x-4) f'(1)=?
已知1+x+x^2+x^3=0,求x+x^2+x^3+x^4+x^5+x^6+x^7+x^8的值