(-x)^3 × x^2n-1 × (-x)^2
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 11:52:48
x)ЭЌ3V8<]"(O$/D?HVBMYX NX$m26Ѵ/.H̳* pE
(-x)^3x^n-1+x^2n(-x)^3
x^n*x^n+1*(-x)^2n*x+(-x)^2n+3x^2n-2*x
(4x^n-2x^n-1-3x^n+2)÷(-5x^n-1)
计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1)
x^n-1(3x^n+4x^n+1-5x^n+2)
C语言 f(x)=1+x+x^2/2!+x^3/3!+...+x^n/n!直到|x^n/n|
x^(n)*x^(n+1)+x^(2n)*x
3x^n(1-x)-2(x^n-2x^n+1) 用提取公因式法
(8x^n+1-3x^n-1)-2(-x+9x^n+1-4x^n)=
已知 x ^3n-2 ÷x^ n+1 =x^3-n×x^n+2,求n的值
分解因式(1+x+x^2+x^3+.x^n)^2-x^n
约分(x^n+2)+(x^n+1)-6x^n/(2x^n+1)+5x^n-(3x^n-1)
( x)^3×(x)^2n 1+(x)^2n×( x)^2(x)^3×(x)^2n-1+(x)^2n×(x)^2
计算:1/(x-1)+1/(x-1)(x-2)+1/(x-2)(x-3)+.+1/(x-n)(x-n-1)
f(x)=x(x-1)(x-2)(x-3).(x-n),则f(x)的n+1阶求导
因式分解:(x^n+1)+(2x^n)+(x^n-1)
x^n+1-2x^n+x^n-1因式分解
x^n-1-2x^n+x^n-1因式分解