已知x²-5x y-1=0 则x y的最大值

1/x-1/y=(y-x)/xy=-3y=x=-3xyx-y=3xy所以原式=[5(x-y)+xy]/(x-y)-xy]=[5(3xy)+xy]/[(3xy)-xy]=16xy/2xy=8

13x^2-6xy+y^2-4x+1=09x²-6xy+y²+4x²-4x+1=0(3x-y)²+(2x-1)²=0∵(3x-y)²≥0&n

（3x-4y+2xy）-（2x-5y+5xy）=x+y-5xy=10

楼主是不是把题写错了呀,应该是（X+2)^2+/Y+1/=0吧因为（X+2)^2大于等于0/Y+1/大于等于0所以只有同时为0时,和才为0即X=-2Y=-1带入原式=4

（x+2）平方+|y+1|=0则x+2=0,y+1=0所以x=-2.y=-1xy平方=-2.x平方y=-45xy*-｛2x*y-【3xy*-（4xy*-2x*y）】｝=5*（-2）-【2*（-4）-3

（x+2）^2+|y+1|=0x=-2,y=-15xy²-｛2xy²-[3xy²（4xy²-2x²y）]｝=5xy²-2xy²+3

根据已知得到x=-2y=-1原式=5xy^2-{2x^2y-[3xy^2-(4xy^2-2x^2y)]}=5xy^2-{2x^2y-[3xy^2-4xy^2+2x^2y]}=5xy^2-{2x^2y-

1/x-1/y=3(y-x)/xy=3y-x=3xyx-y=-3xy(5x+xy-5y)/(x-xy-y)=[5(x-y)+xy][(x-y)-xy]=(-15xy+xy)/(-3xy-xy)=-14

x^2y+xy^2=xy(x+y)=1/5

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

(x+1)平方+y-1的绝对值=0x+1=0y-1=0x=-1y=12（xy-5xy平方）-（3xy平方-xy）=2xy-10xy^2-3xy^2+xy=3xy-13xy^2=-3+13=10

(x+1)^2+|y-1|=0(x+1)^2>=0|y-1|>=0x+1=0,y-1=0x=-1,y=12(xy-5xy的平方)-(3xy的平方-xy)=2xy-10xy^2-3xy^2+xy=3xy

（x+1)^2+|y-1|=0x+1=0x=-1y-1=0y=12（xy-5xy^2)-(3xy^2-xy)=2xy-10xy²-3xy²+xy=3xy-13xy²=3x

xy+1/xy>=2√（xy*1/xy）=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√（x/y*y/x）=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

x²-3xy+4x=3,y²+xy-4y+7=0相加得x²-2xy+y²+4（x-y）+4=0（x-y）²+4（x-y）+4=0（x-y+2）=0,得

(x+y)^2>=0,(y-1)^2>=0,因此y=1,x=-1.原式=2(-1-5)-(-3+1)=-10再问：打错了是已知(x+1)的平方+(y-1)的绝对值=0,求2(xy-5xy的平方）-（3

13x^2-6xy+y^2-4x+1=09x^2-6xy+y^2+4x^2-4x+1=0(3x-y)^2+(2x-1)^2=03x-y=0,2x-1=0,x=0.5y=3*0.5=1.5(xy-x^2

(x+1)²+|y-1|=0两个非负数的和为0,这两个非负数都为0x+1=0,y-1=0x=-1,y=12(xy-5xy²)-(3xy²-xy)=2xy-10xy

因为,x-y=3xy所以：-3x+6xy+3y/[(7x-5y)-9xy-(3x-y)]=[-3(x-y)+6xy]/[(7x-5y)-9xy-(3x-y)]=[-3*3xy+6xy]/[7x-5y-