x>0,y>0,x y=1,求x2 y2 根号(xy)最值
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由|X-2|+(Y+1/4)^2=0绝对值大于等于0平方也是大于等于0所以X=2Y=-1/4带入后面要求的式子得到等于0-17/8
3x*+xy-2y*=0(3x-2y)(x+y)=0那么x=2y/3或x=-yy/x=3/2或x/y=-1(x/y)-(y/x)-(x*+y*)/(xy)=(x/y)-(y/x)-(x/y+y/x)=
绝对值大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所以x+y+1=0xy+3=0xy=-3x+y=-1两边平方x^2+2xy+y^2=(-1)^2x^2+y^2=1-
即x=y=1xy=1对不对?如果对的话x^2+2y^2-2xy-2y+1=0化简为你做的很对.就是这样解的,没有其他更好的方法了.这里用到的是数学里的
x²+y²-xy+2x-y+1=0x²+2x+1-y(x+1)+y²=0(x+1)²-y(x+1)+y²=0(x+1-y/2)²+
x=4,y=0.5,x+y=4.5(与人家的做法一样……)(1)解题思路是以S3为基准,用S3表示出S1,S2,S4即可.在三角形BCD中有:S2/S3=DF/CF,故S2=(DF/CF)S3;同理,
y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5
X²+Y²+X²Y²-4XY+1=0(X-Y)^2+(XY-1)^2=0所以X-Y=0XY-1=0(X-Y)^2008-(XY)^2008=0^2008-1^2
因为x+2y=0,所以x=-2y原式=(x^2+2xy)/(xy+y^2)=(4y^2-4Y^2)/(-3y^2+y^2)=0/(-2y^2)又因为xy不等于零,所以x、y君不等于零,所以-2y^2亦
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
(x+2)²+|y+1|=0则x+2=0y+1=0x=-2y=-1(2x²-xy)-2(x²-y²+xy)+(2xy-2y²)=2x²-xy
y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5
x²+y²-xy+2x-y+1=[3(x+1)²+(x-2y+1)²]/4=0,由于(x+1)²>=0且(x-2y+1)²>=0,则有x+1
3-y=0x+y=0∴x=-3y=3[2(x+y)-3(xy+4)]÷1/xy=[2×0-3(-9+4)]÷(-1/9)=15×(-9)=-135
10x²-2xy+y²+6x+1=0(3x+1)²+(x-y)²=03x+1=0x-y=0所以x=y=-1/3x+y=-2/3再问:3x+1=x-y=再答:3x
2x^2+2xy+y^2-2x+1=0所以x^2+2xy+y^2+x^2-2x+1=0所以(x+y)^2+(x-1)^2=0所以x+y=0,x-1=0所以x=1,y=-1所以xy=-1
5x²+5xy+y²+2x+1=04x²+y²+4xy+2x+x²+1=0(2x+y)²+(x+1)²=0得{2x+y=0---①
①xy同非负时,2x-3√xy-2y=(2√x+√y)(√x-2√y)=0∴√x=2√y,x=4y②xy同负时,2x-3√xy-2y=[2√(-x)+√(-y)][√(-x)-2√(-y)]=0∴√(
x+y=5xy(2x-3xy+2y)/(x+2xy+y)=[2(x+y)-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问:若x+1/x=3,求(x
0=x^2+y^2+x^2y^2-4xy+1=x^2+y^2-2xy+x^2y^2-2xy+1=(x-y)^2+(xy-1)^2,两个平方数的和等于0,所以,x=y,xy=1,带入得(x-y)^200