设实数x,y满足方程9x2 4y2-3x 2y=0
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x^2+y^2-8x-6y+21=0配方,有:x^2-8x+16+y^2-6y+9=4(x-4)^2+(y-3)^2=4这是一个以(4,3)为圆心,2为半径的圆令x-4=2sinay-3=2cosa(
令2x+y=py=p-2x3x^2+2y^2-6=3x^2+2(p-2x)^2-6=11x^2-8px+2p^2-6≤0△=64p^2-4*11(2p^2-6)=-24p^2+24*11≥0p^2≤1
答题过程:9X^2+4Y^2-3X+2Y=(3x-1/2)^2+(2y+1/2)^2-1/2=0-->(3x-1/2)^2+(2y+1/2)^2=1/2设3x-1/2=a,2y+1/2=b则有:a*a
【解】设a=xy²,b=x²/y.(x³)/(y^4)=b²/a由题设可得:①3≦a≦8.∴1/8≦1/a≦1/3.②4≦b≦9.∴16≦b²≦81.
x+y=ay=a-x代入2x²+3a²-6ax+3x²-6a+6x=05x²+(6-6a)x+(3a²-6a)=0x是实数所以△>=036-72a+3
你的题不全,原题为:设实数X,Y满足(X的平方+Y的平方乘4+2X-4Y+2=0,求X的2Y次方+Y的开方乘2X的值等于多少?(x+1)^2+(2y-1)^2=0所以x=-1,y=1/2代入x^2y+
设x=√2sinθ,y=√3cosθ得p=2x+3y=2√2sinθ+3√3cosθ=√35sin(θ+φ)最大值为根号下35
条件:0
9用线性规划就行了
∵(x2+y2)2=x4+y4+2x2y2,而x4+y4=72,设x2+y2=t>0,∴t2=2x2y2+72,又∵x+y=1,∴(x,+y)2=x2+2xy+y2=1,∴xy=1−t2,∴t2=2•
(1/2+π/3)X+(1/3+π/2)Y=π+4(2π+3)x+(3π+2)y=6n+242πx+3πy+3x+2y=6n+24{2x+3y=6(1)3x+2y=24(2)(1)*3得6x+9y=1
B.0
解由2x2+3y2=4x得2x2-4x+3y2=0即2(x-1)^2+3y^2=2即(x-1)^2+y^2/(2/3)=1故由三角函数知识设x=1+cosa,y=√6sina/3则x+y=1+cosa
令x+y=k,则k-x=y,代入原式,有2x^2+3(k-x)^2=4x,化简得5x^2-(4+6k)x+3k^2=0,其判别式大于等于0,则(4+6k)^2-60k^2>=0,然后最终可以得到关于k
满足约束条件的平面区域如下图所示:联立x=yx+2y=3可得x=1y=1.即A(1,1)由图可知:当过点A(1,1)时,2x-y取最大值1.故答案为:1
2y=z-3x所以9x²+z²-6xz+9x²-3x+z-3x=018x²-(6z+6)x+z²+z=0x是实数所以△>=036z²+72z
x+y
设x^3/y^4=(xy^2)^m*(x^2/y)^n则:3=m+2n-4=2m-n解得:m=-1,n=2所以x^3/y^4=(x^2/y)^2/(xy^2)因为4
令t=2x+y,可得y=t-2x,代入x2+y24=1,得x2+14(t-2x)2=1化简整理,得2x2-tx+14t2-1=0∵方程2x2-tx+14t2-1=0有实数根∴△=t2-4×2×(14t
∵正实数x,y,z满足x+2y+z=1,∴1x+y+9(x+y)y+z=x+y+y+zx+y+9(x+y)y+z=1+y+zx+y+9(x+y)y+z≥1+2y+zx+y×9(x+y)y+z=7,当且