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x^3+y^3=16 x^3+y^3=(x+y)(x^2-xy+y^2) =(x+y)[(x+y)^2-3xy] x+y>=2√xy (x+y)^2/4>=xy (x+y)^2-3xy=4

解:∵y''(e^x+1)+y'=0 ==>(e^x+1)dy'/dx=-y' ==>dy'/y'=-dx/(e^x+1) ==>dy'/y'=-e^(-x)dx/(1+e^(-x)) ==>dy'/y'=d(1+e^(-x))/(1+e^(-x)) ==>ln│y'│=ln(1+e^(-x))+ln│C1│ (C1是积分常数) ==>y'=C1(1+e^(-x)) ==>y=C1(x-e^(-x))+C2 (C1是积分常数)...

∵(x+y)dx+(y-x)dy=0 ==>(1+y/x)dx+(y/x-1)dy=0 设y=xt,则dy=tdx+xdt ∴(x+y)dy+(y-x)dx=0 ==>(1+t)dx+(t-1)(tdx+xdt)=0 ==>(t²+1)dx+x(t-1)dt=0 ==>dx/x+(t-1)/(t²+1)dt=0 ==>ln|x|+∫t/(t²+1)dt-∫1/(t²+1)dt=ln|C1| (C是积分...

解: (1) x+2y=3k ① 2x+y=-2k+1 ② ①+② 3x+3y=k+1 x+y=(k+1)/3 x+y>3/5 (k+1)/3>3/5 k+1>9/5 k>4/5 k的取值范围为(4/5,+∞) (2) k>4/5,5k+1>0,4-5k

设x1>x2 f(x1)-f(x2) =f[(x1-x2)+x2]-f(x2) =[f(x1-x2)+f(x2)]-f(x2) =f(x1-x2) 因为x1>x2 所以x1-x2>0 所以f(x1-x2)

1/y=(x^2+x+1)/(x+2)=(x^2+4x+4-3x-6+3)/(x+2) =(x+2)-3+3/(x+2) 因为x>-2 所以x+2>0 (x+2)+3/(x+2)>=2根号3 所以1/y>=2根号3-3 当1/y取最小值时,y有最大值 此时x+2=3/x+2 (x+2)^2=3 x^2+4x+1=0 因为x>-2 得x=-2+根号3 此时y=1/(2根号3-3)=(2...

如果要输入x和y的值应该是cin>>x>>y; 如果是cin>>x+y, x+y会被先执行, 它执行完后是x的值和y的值相加, 这样就是一个常量了.

已知x、y、z都是非负实数,且x+y+z=1 由于对称性,不妨假设1>=x>=1/3>=y>=z>=0,则yz=8/18=4/9,则f(x,y,z)=2(2x-1)^2-8yz+18yzx>=2(2x-1)^2>=0 否则1/3=2/9,5/18

#include int main(void) { int x,y; printf("pls input the x = %d"); scanf("%d", &x); //阶梯if if (x =0) y = x+1; else y = -(x+1); } else if (x < 2) y = x - 1; else y = 2x; //嵌套 if (x = 0) y = x+1; if (x -1 && x < 2) y = x -1; ...

解:原式=∫dθ∫((cosθ+sinθ)r+1)rdr (作变换:x=rcosθ+1/2,y=rsinθ+1/2) =∫dθ∫((cosθ+sinθ)r^2+r)dr =∫((√2/12)(cosθ+sinθ)+1/4)dθ =π/2。

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