已知x-y 2大于等于0 x y-4大于等于0 2x-y-5小于等于0
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∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,(x-1)2+(y-2)2=0,∴x=1,y=2,∴yx−xy=2-12=1.5;故答案为:1.5.
u=x2+y2+4x-2y=(x+2)^2+(y-1)^2-5>=-5所以(x+2)^2+(y-1)^2=u+5>0这里可以看作是以(-2,1)为圆心,半径为根号(u+5)的圆(只是这里的圆半径会变,
x²-3xy-4y²=0x²+4xy+4y²-1=0由(1)得(x-4y)(x+y)=0x=4yx=-y由(2)得(x+2y)²-1=0x+2y=1x
∵x2-5xy+6y2=0,∴(x-2y)(x-3y)=0∴x-2y=0,x-3y=0,即x=2y,x=3y,∴y:x等于12或13.故选C.
∵x2+y2-4x+6y+13=(x-2)2+(y+3)2=0,∴x-2=0,y+3=0,即x=2,y=-3,则原式=(x-3y)2=112=121.
已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?2x-3*根号(xy)-2y=0(根号X-2根号Y)(2根号X+根号Y)=0根号X-2根号Y
X2+Y2+8X+6Y+25=0x²+8x+16+y²+6y+9=0(x+4)²+(y+3)²=0∴x+4=0y+3=0x=-4y=-3X2+4XY+4Y2分之
(x2y)=(x2y)(8/x1/y)=8216y/xx/y
把题写纸上传个图片吧,这样题目有歧义啊
原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y,由已知得(3x-2y)(x+y)=0,因为x+y≠0,所以3x
最大值为八分之一再问:确定?再答:嗯
x大于等于0,y大于等于0,且x+2y=1,0
xy+1/xy+y/x+x/y=[(xy)^2+1+x^2+y^2]/(xy)=[(xy)^2-2xy+1+x^2-2xy+y^2+4xy]/(xy)=[(xy-1)^2+(x-y)^2+4xy]/(
∵x2-4xy+4y2=0,∴(x-2y)2=0,∴x=2y,∴x-yx+y=2y-y2y+y=13.故分式x-yx+y的值等于13.
(x+y)^2=x∧2+y∧2+2xy=9x∧2+y∧2=9-2×(-9)=27
已知x大于0,Y大于0,XY等于8,所以:f(x,y)=x+2y=x+2*8/x=x+16/xdf(x,y)/dx=1-16/(x^2)==>当df(x,y)/dx=1-16/(x^2)=0,即x=4
xy≦(x^2+y^2)/2,当x=y时等号成立,这时xy取最大值;因为x+y=4,所以当x=y时,x=y=2,所以xy的最大值为(x^2+y^2)/2=4.再问:其他方法呢?再答:(1)x+y=4,
6x2-xy-15y2=(2x+3y)(3x-5y)=0,所以x=-3/2y或x=5/3y
设k=9x-y,则y=9x-k,代入已知式,得-4≤x-(9x-k)≤-1,-1≤4-(9x-k)≤5(改题了),即8x-4≤k≤8x-1,9x-5≤k≤9x+1,画示意图知,由k=8x-4,k=9x
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²