k^2>16/25 ,m=4(25k^2+9)/(-25),求m的取值范围
来源:学生作业帮助网 编辑:六六作业网 时间:2025/01/24 14:44:49
k^2>16/25,m=4(25k^2+9)/(-25),求m的取值范围k^2>16/25,m=4(25k^2+9)/(-25),求m的取值范围k^2>16/25,m=4(25k^2+9)/(-25)
k^2>16/25 ,m=4(25k^2+9)/(-25),求m的取值范围
k^2>16/25 ,m=4(25k^2+9)/(-25),求m的取值范围
k^2>16/25 ,m=4(25k^2+9)/(-25),求m的取值范围
25k²>16
25k²+9>25
(25k²+9)/(-25)<-1
m<-1
k^2>16/25 ,m=4(25k^2+9)/(-25),求m的取值范围
(k*k*k-2k+4)/4k
设向量m=(1-k,1-k,k),n=(2,k,k),则|m-n|的最小值
4k+m/4=-3,k+m/2=3k和m分别为多少
|-3k-1+n-km|/√(k^2+1)=|-4/k-5+n+m/k|/√(1/k^2+1)这个怎么化简
化简|-3k-1+n-km|/√(k^2+1)=|-4/k-5+n+m/k|/√(1/k^2+1)
4^2k-4^k=2,k=?
k^4+2k^2-6k-3=0
3k-4k=2 k=?
main(){int j,k,s,m;for( k=1 ; k < 10; k++ ){s=1; m= k+2;for( j=k; j
1/2(m-4)^2+k=01/2(m-2)^2+k=-2
#define SQR(X) X*X #include void main(){ int a=16,k=2,b=4,m=1 ; a/=SQR(k+m)/SQR(k+m); pri#define SQR(X) X*X#include void main(){int a=16,k=2,b=4,m=1 ;a/=SQR(k+m)/SQR(k+m); printf(%d
,a); }为什么?
k^2+k-4=2咋算?
求证:lim1^k+2^k+3^k+4^k+.n^k/n^(k+1)=1/k+1n是正整数,后面的k+1有括号的
无论k取何值,直线3(k+2)x+(5k-1)y-(4k-3)=0横过一个定点M,求点M的坐标?
p(k)=2m(m+1) / [k(k+1)(k+2)];这个公式中( )里面的数值,
x1=@sqrt((m-m1)*(m-m1)+k*k*m*m-2*(m-m1)*k*m*@sin(a))-(m1-m(1-k));这个lingo语句哪里错误了?
[k*(2-4k)/(1+2k)]+2k+1