1.(16x^2+1)(4x^2+1)(2x+1)(1-2x)2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+11.(16x^4+1)(4x^2+1)(2x+1)(1-2x)第一题是这样的……抱歉2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1第二题是这样的……抱歉

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/18 00:48:17
1.(16x^2+1)(4x^2+1)(2x+1)(1-2x)2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+11.(16x^4+1)(4x^2+1)(2x+1)(1-2x)第一题是这样的……抱歉2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1第二题是这样的……抱歉

1.(16x^2+1)(4x^2+1)(2x+1)(1-2x)2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+11.(16x^4+1)(4x^2+1)(2x+1)(1-2x)第一题是这样的……抱歉2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1第二题是这样的……抱歉
1.(16x^2+1)(4x^2+1)(2x+1)(1-2x)
2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
1.(16x^4+1)(4x^2+1)(2x+1)(1-2x)
第一题是这样的……抱歉
2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
第二题是这样的……抱歉

1.(16x^2+1)(4x^2+1)(2x+1)(1-2x)2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+11.(16x^4+1)(4x^2+1)(2x+1)(1-2x)第一题是这样的……抱歉2.(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1第二题是这样的……抱歉
1
(16x^4+1)(4x^2+1)(2x+1)(1-2x)=(1+16x^4)(1+4x^2)[(1+2x)(1-2x)]=(1+16x^2)[(1+4x^2)(1-4x^2)]
=(1+16x^4)(1-16x^4)=)=1-256x^4
2
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1=(2^4-1)(2^4+1)(2^8+1)(2^16+1)+1
=(2^8-1)(2^8+1)(2^16+1)+1=(2^16-1)(2^16+1)+1=(2^32-1)+1
=2^32
3
9×11×101×10001=(10-1)(10+1)(10^2+1)(10^4+1)
=(10^2-1)(10^2+1)(10^4+1)=(10^4-1)(10^4+1)=10^8+1=100000001

*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || 1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做 1.用公式计算:COSX=1-X*X/2!X*X*X*X/4!-X*X*X*X*X*X/6!.直到最后一项的绝对 (x+2)(x-2)+16 (x-1)(x-2)+3x(x+3)-4(x+2)(x-3)(x+2)(x-2)+16 (x-1)(x-2)+3x(x+3)-4(x+2)(x-3) 计算(x+/1x)(x^2+1/x^2)(x^4+1/x^4)(x^8+1/x^8)(x^16+1/x^16)(x^2-1) (x+1/x)(x^2+1/x^2)(x^4+1/x^4)(x^8+1/x^8)(x^16+1/x^16)(x^2-1) (x+x^-1)(x^2+x^-2)(x^4+x^-4)(x^8+x^-8)(x^16+x^-16)(x^2-1)=? 已知x=2π+1.求(X-1)+(X-2)+(X-3)+(X-4)+(X-5)+(X-6)+(X-7)+(X-8)+(X-9)+(X-10) (x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x-1) x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4 解方程:2x(x-2)-6x(x-1)=4x(1-x)+16 1.(2x-1)(4x²+2x+1))2.(x-1)(x²+x) x=1-x/2+x/4-x/8+x/16-x/32 x^5-x^4+x^3-x^2+x-1 x^6-x^5+x^4-x^2+x-1 求一个数独答案X X X 9 X X X 8 2 X 6 3 X X 1 4 X 99 X 8 X X X X X XX X X 6 7 X 3 X XX 4 6 X 5 X 2 9 XX X 7 X 2 3 X X XX X X X X X 7 X 17 X 4 3 X X 6 2 X6 3 X X X 7 X X X 一个“整式的乘法”的问题请先阅读下列解题过程,再仿做下面的问题.已知X*X + X - 1=0,求X*X*X + 2*X*X + 3的值.解X*X*X + 2X*X +3=X*X*X +X*X -X +X*X +X +3=X{X*X +X -1} +X*X +X -1 +4=0+0+4=4+ x + X*X + X*X*X=0.+ X*X + X*X*X 24道符号^表示平方1,x^+21x+202,x^-8x-203,x^+8x-204,x^-8x-335,x^-4x-326,x^-14x+337,x^+49x+488,x^-14x+489,x^-2x-4810,x^+22x-4811,x^-19x+4812,x^-16x+4813,x^+8x-4814,x^-29x+10015,x^+x-7216,x^+32x-3317,x^-21x-10018,x^+4x+319,x^-6x+520,x^+15x+2621,x^