x>0 y>0 x+y=1 证(x+1/x)(y+1/y)≥25/4
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/20 02:58:24
x>0y>0x+y=1证(x+1/x)(y+1/y)≥25/4x>0y>0x+y=1证(x+1/x)(y+1/y)≥25/4x>0y>0x+y=1证(x+1/x)(y+1/y)≥25/4(x+1/x)
x>0 y>0 x+y=1 证(x+1/x)(y+1/y)≥25/4
x>0 y>0 x+y=1 证(x+1/x)(y+1/y)≥25/4
x>0 y>0 x+y=1 证(x+1/x)(y+1/y)≥25/4
(x+1/x)(y+1/y)
=xy+x/y+y/x+1/(xy)
=(x/y+y/x)+xy+1/(xy)]
依均值不等式x/y+y/x>=2.(*)
因为xy=0时,在X=1时有极小值Y=2,在X>1时递增,在0=2+17/4=25/4.
x/y=y/x,xy=1/4--->x=y=1/2,时“=”成立.
(x+1/x)(y+1/y)=(1+1/x)(1+1/y)
然后用均值不等式...
最后用x+y=1 检验。
应该会了吧?
X>0 Y>0证1/x+1/y>=4/(x+y)X>0 Y>0 证1/x+1/y>=4/(x+y)
设x大于1,y大于0,x^y+x^-y=2根号二,x^y-x^-y等于?
x>0 y>0 x+y=1 证(x+1/x)(y+1/y)≥25/4
mathematica软件,已知y[x],Y=f1(x),X(x)=f2(x),如何plot Y[X]?y[x_] := x + 9.81/2*x*xk=1X[x] = x - k*y'[x]/(1 + (y'[x])^2)^0.5Y[x] = y[x] + k/(1 + (y'[x])^2)^0.5Plot[{y[x],Y[X],},{x,0,2}]
若实数x,y满足x-y+1》=0,x+y>=0,x
已知x,y满足约束条件:x-y+1>=0,x+y-2>=0,x
若实数x,y满足x-y+1>=0,x+y>=0,x
画出不等式组x+y>0,x=还有不等式(x-y)(x-y-1)=
实数x,y满足{x-y+1>=0;y+1>=0;x+y+1
x>0 y>0 x+y=1,4/X+9/y最小值x>0 y>0 x+y=1,求4/X+9/y最小值
y=1/x^2(x>0) y=(1-x)/(1+x) y=(e^x-e^-x)/2
y=1-x^2 x>=0 ,y=sin|x|/x x
若x+y=1,x不等于0则x+(2xy+y^2)/x/(x+y)/x=
若|x+y-1|+(x-y-2)²=0,求代数式(x+2y)(x-2y)-(2x-y)(-y-2x)的值.
已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)
已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)
已知x²+y²+6x-8x+25=0,求(x/x²-y² - 1/x+y)÷y/y-x
y=x(1-x) (0