limcotx[√(X 1)-1] x
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(x1+1/x1)-(x2+1/x2)=(x1-x2)+(1/x1-1/x2)第一步:(x1+1/x1)-(x2+1/x2)去括号得x1+1/x1-x2-1/x2第二步:x1+1/x1-x2-1/x2
x1x2..xn均为整数应是x1x2..xn均为正数吧,由均值不等式得:(x2/√x1)+√x1≥2√x2,(x3/√x2)+√x2≥2√x3,...(x1/√xn)+√xn≥2√x1,把上面n个不等
1.1x1.1x1.1-1.1x1.1-0.11=1.1x1.1x1.1-1.1x1.1x1-0.11x1=1.1x1.1x(1.1-1)-0.11x1=1.1x1.1x0.1-0.11x1=1.1x
1x1/2+1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/9-1/7=1-1/7=6/
当m=1时,B={x|13==>m>3综上,符合条件的实数m的取值范围是m3再问:是求A∪B,不是求A∩B全集C是R再答:(1)当m=1时,A={x|-1<x≤3},B={x|1≤x<4},则A∪B=
一成不变
再答:根与系数的关系再答:不懂请追问再答:
和高手讨论了一下,这办法不是我想的.(x1/(1+x1^2)+x2/(1+x1^2+x2^2)+...+xn/(1+x1^2+x2^2+...+xn^2))^2
model:min=45/(x1+x2);x1+x2>=1;x1
1.1x1.1x1.1-1.1x1.1-0.1=1.1×1.1×(1.1-1)=1.1×1.1×0.1-0.1=0.121-0.1=0.021
x1=3/2;x2=x1/2;printf("%f\n",x1);你会发现x1就等于1因为x1=3/2;3和2都是整型,除下来结果也为整型,是1,然后赋值给float,变成1.0
1x1\3=1/2*(1/1-1/3)2x1\4=1/2*(1/2-1/4).1x1\3+2x1\4+3x1\5+.+2006x1\2008=1/2(1/1-1/3+1/2-1/4+1/3-1/5+.
和高手讨论了一下,这办法不是我想的.(x1/(1+x1^2)+x2/(1+x1^2+x2^2)+...+xn/(1+x1^2+x2^2+...+xn^2))^2
提取公因式(x1-x2)
因为1/[n(n+2)]=1/2*[1/n-1/(n+2)]所以1/4x1/6+1/6x1/8+1/8x1/10+1/10x1/12=1/2*(1/4-1/6+1/6-1/8+1/8-1/10+1/1
Xn/(x1+x2+...Xn-1)(X1+X2...+Xn)=1/(x1+x2+...+xn-1)-1/(x1+x2+...+xn-1+xn)所以原式=1/x1-1/(x1+x2)+1/(x1+x2
length(x1)%返回x1的长度,zeros(1,y)%返回一个1行y列的向量,数值全为0.x1=[x1zeros(1,N-length(x1))];%在x1向量后面补充0,使其长度变为N.
提取公因式(x1-x2)原式=(x1-x2)]1-4/x1x2]
=1.1×1.1×(1.1-1)-0.11=1.1×1.1×0.1-0.11=0.11×(1.1-1)=0.11×0.1=0.011