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x2y x3 y3

沿路径y=x^(3/2),趋向于无穷,limxy/(x^3+y^2)=lim1/2x^(1/2)->∞因此极限不存在。这种题无非两种,一种取路径,一种极坐标。

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.

x3+x2-y3-y2 =x3-y3+x2-y2 =(x-y)(x2+xy+y2)+(x-y)(x+y) =(x-y)(x+y+x2+xy+y2)

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

f'x=3x²-3=0, 得x=-1 , 1 f'y=-3y²+12=-3(y²-4)=0, 得y=-2, 2 则有4个驻点(1, -2), (1, 2), (-1, -2), (-1, 2) A=f"xx=6x B=f"xy=0 C=f"yy=-6y B²-AC=36xy, 因此当x,y同号时,B²-AC>0, 不是极值点;当x,y异号时,才是...

(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...

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...

#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...

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-...

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