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x2y x3 y3

若y=kx²,lim=lim(1+k)/(x+k³x^4)=∞ 若y=-x²,lim=lim0/(x³-x^6)=0 所以极限不存在

分子趋向于0,分母也趋向于0,不能直接判断

证明:(1)∵(x3+y3 )-(x2y+xy2)=x2 (x-y)+y2(y-x)=(x-y)(x2-y2 ) =(x+y)(x-y)2.∵x,y都是正实数,∴(x-y)2≥0,(x+y)>0,∴(x+y)(x-y)2≥0,∴x3+y3≥x2y+xy2.(2)函数f(x)=|x+1|+|x-5|,x∈R,关于x的不等式f(x)≥a-...

5求方程 X3+Y3-X2Y2-(X+Y)Z=0的所有非负数整数解: Z = (X^3+Y^3-X^2-Y^2)/(X+Y) (1) Z = {X^2(X-1) + Y^2(Y-1)}/(X+Y) (2) X(X-1) 其中:X+Y> 0 X,Y不同时为1 (3) 有无穷多组解:{X,Y,Z}={n,n,n(n-1)},(n>=2) {2,2,2},{3,3,6},{4,4,12},{5,5,2...

#include #include int main() { int *x1,*y1,*x2,*y2,*x3,*y3,T,i; scanf("%d",&T); x1=(int *)malloc(sizeof(int)*T); y1=(int *)malloc(sizeof(int)*T); x2=(int *)malloc(sizeof(int)*T); y2=(int *)malloc(sizeof(int)*T); x3=(int *)mallo...

#include #include float dis(float x1,float y1,float x2,float y2){float dx,dy;dx=x1-x2;dy=y1-y2;return sqrt(dx*dx+dy*dy);}int main(){float x1,x2,x3,y1,y2,y3,p,s,a,b,c;scanf("%f %f",&x1,&y1);scanf("%f %f",&x2,&y2);scanf("%f %f",&...

x3+y3=(x+y)(x2-xy+y2),∴x2-xy+y2=10,∵x+y=10,∴x2+2xy+y2=100,∴2xy=100-(x2+y2),把xy=x2+y2-10,代入得:100-(x2+y2)=2(x2+y2-10)=2(x2+y2)-20,3(x2+y2)=120,∴x2+y2=40.故答案为:40.

(1)证明:∵(x3+y3)-(x2y+xy2)=x2(x-y)+y2(y-x)=(x-y)(x2-y2)=(x-y)2(x+y)又x、y都是正实数,∴(x-y)2≥0、x+y>0,即(x3+y3)-(x2y+xy2)≥0∴x3+y3≥x2y+xy2;(2)∵a3-b3=a2-b2,∴(a-b)(a2+ab+b2)=(a-b)(a+b),又a≠b...

当极限沿x=-y趋于0的时候,极限是不存在的,因为分母无意义 所以在一个方向上的极限不存在,那么二重极限不存在

A,B,C三点共线 (y₂-y₁)/(x₂-x₁)=(y₃-y₂)/(x₃-x₂) 即(x₂³-x₁³)/(x₂-x₁)=(x₃³-y₂³)/(x₃-x₂) (x₂-x₁)(xS...

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