x^2-x-2n

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x^2-x-2n
(x+y)(m+n)2-(x+y)(m+n)2

+Y=M+N反应前后质量保持不变,所以5gX和3gY完全反应,除生成1gM,其余的都是N,即生成N7g要制取14gN,即要两倍以上质量的反应物反应.即要10gX和6gY反应.根据质量首恒定律可知N=X

\求和Sn=1+2x+3x^2+```+(n-1)x^(n-2)+n*x^(n-1)

若x=1Sn=1+2+3+……+n=n(n+1)/2若x不等于1xSn=x+2x^2+3x^3+……+n*x^n所以Sn-x*Sn=1+x+x^2+x^3+……+x^(n-1)-n*x^nSn(1-x

f(x)=x(x-1)(x-2)(x-3).(x-n),则f(x)的n+1阶求导

f(x)为n+1阶多项式,所以n+1阶求导后只会剩下x的n+1次方的导数,为n+1的阶乘

单项式乘以多项式(x^n-x^n-1+x)*x^n+1(x^2n+1)(-x^2n)(+x^n+2)

[x^n-x^(n-1)+x]*x^(n+1)=x^(2n+1)-x^2n+x^(n+2)

若(x^2)^n=x^10则n等于?

(x^2)^n=(x)^2n=x^10,则2n=10,n=5

x^2n(x^2-2n-2x^1-2n+x^-2n)

答案是(x-1)^2注意,2x^3y指的是2*x^3*y;你的意思,正确的写法是2x^(3y)再问:额就写答案鬼知道你步骤是什么?再答:这个题目直接:x^(2n)*x^(2-2n)-x^(2n)*x^

C语言 f(x)=1+x+x^2/2!+x^3/3!+...+x^n/n!直到|x^n/n|

#include<stdio.h>#include<math.h>//f(x)=1+x+x^2/2!+x^3/3!+...+x^n/n!直到|x^n/n|<10^-6do

{x|x=2n,n∈Z}文字表述

在xy坐标中有一个坐标点(x,y),满足x+y=0,x-y=0在xy坐标中满足y=x^2且x属于全集

计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1)

原式=x^(5n)-x^(4n)+x^(2n)+x^(4n)-x^(3n)+x^n+x^(3n)-x^(2n)+1=x^(5n)+x^n+1

计算(-x)^2n十1.(-x)^n+1

(-x)^2n十1.(-x)^n+1=(-x)^3n+2=-x^3n+2

若x^m=6,x^n=9,则(2x^3m*x^2n) / (x^m*x^n)^2*x^n

(2x^3m*x^2n)/(x^m*x^n)^2*x^n=(2*6^3*9^2)/(6*9)^2*9=4/3

计算:(m+2n)/(n-m)+n/(m-n)-2m(n-m)和[(x+2)/(x×x-2x)-(x-1)/(x×x-4

(m+2n)/(n-m)+n/(m-n)-2m(n-m)=(m+2n-n-2m)/(n-m)=(n-m)/(n-m)=1[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/

因式分解4x^(n+2)-9x^n+6x^(n-1)-x^(n-2)

4x^(n+2)-9x^n+6x^(n-1)-x^(n-2)=x^(n-2)(4x^4-9x^2+6x^2-1)=x^(n-2)[4x^4-(3x-1)²]=x^(n-2)(2x²

若m-2n,x

n-2n和x

求极限lim [x^(n+1)-(n+1)x+n]/(x-1)^2 x趋于1

lim(x->1)(x^(n+1)-(n+1)x+n)/(x-1)^2=lim(x->1)(x^(n+1)-(n+1)x+n)'/((x-1)^2)'=lim(x->1)((n+1)x^n-(n+1)

计算x^3n/(x^2-1)-x^2n/(x^n+1)-1/(x^n-1)+1/(x^n+1)

x^3n/(x^n-1)-x^2n/(x^n+1)-1/(x^n-1)+1/(x^n+1)=(x³ⁿ-1)/(xⁿ-1)-(x²ⁿ-1)/(x&

求不等式x/2+x/6+x/12+x/20+...+x/(n-1)n>n-1的解集

x/2+x/6+x/12+x/20+...+x/(n-1)n>n-1x×[1/2+1/6+1/12+1/20+……+1/(n-1)n]>n-1x×[1-1/2+1/2-1/3+1/3-1/4+1/4-

数学x^n*x^n+1+x^2n*x在线等.

x^n*x^n+1+x^2n*x=x^(2n+1)+x^(2n+1)=2x^(2n+1)