1=x² 4y²-2xy,则x 2y取值范围

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1=x² 4y²-2xy,则x 2y取值范围
x(x-1)-(x2-y)=-3,求x2-y2-2xy的值

解题思路:由完全平方公式可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/rea

已知x2+y2-2x-4y+5=0,则1xy+1(x+1)(y+1)+1(x+2)(y+2)+…+1(x+2010)(y

∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,∴(x-1)2+(y-2)2=0,∴x=1,y=2,∴原式=11×2+12×3+…+12011×2012=1-12+12-13+

已知:x-2y=12,xy=-25,求x2+4y2-1的值?

x-2y=12,两边平方得,x2+4y2-4xy=144,移项得,x2+4y2-1=4xy+143,将xy=-25代入上式得,x2+4y2-1=43.答:x2+4y2-1的值是:43.

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?

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

已知x+y=4,xy=2,则x2+y2+3xy=______.

x2+y2+3xy=(x+y)2+xy,∵x+y=4,xy=2,∴x2+y2+3xy=42+2=18.故答案为18.

若实数XY满足X2+Y2=1,则X-2Y的最大值为

设:S=x-2y,x=S+2y代入x²+y²=1中得:(s+2y)²+y²=15y²+4sy+s²-1=0∵y是实数,∴△≥0(4s)&su

实数x,y满足x≥y≥1和2x2-xy-5x+y+4=0,则x+y=______.

∵2x2-xy-5x+y+4=0∴x2+x2-xy-4x-x+y+4=0∴x2-4x+4+x(x-y)-(x-y)=0∴(x-2)2+(x-y)(x-1)=0∵(x-2)2≥0,x≥y≥1,∴(x-y

设x,y为实数,若4x2+y2+xy=1,则2x+y的最大值是(  )

∵4x2+y2+xy=1∴(2x+y)2-3xy=1令t=2x+y则y=t-2x∴t2-3(t-2x)x=1即6x2-3tx+t2-1=0∴△=9t2-24(t2-1)=-15t2+24≥0解得−21

若x<y<0,则x2−2xy+y2+x2+2xy+y2=(  )

∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x

实数xy满足x2+y2+2x-4y+1=0,则x2+y2-2x+1的最小值为

x^2+y^2+2x-4y+1=0(x+1)^2+(y-2)^2=2^2x=-1+2cosx,y=2+2sinxy-x=3+2(sinx-cosx)=3+2√2(sinx/√2-cosx/√2)=3+

x(x-1)-(x2-y)=-3,求x2+y2-2xy的值

x(x-1)-(x2-y)=-3x2-x-x2+y=-3-x+y=-3所以x-y=3所以原式=(x-y)2=9

已知x(x-1)-(x2-y)=-3,求x2+y2-2xy的值

解题思路:完全平方公式解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.

已知x+y=4,xy=2,求x2+y2+3xy的值.

∵x+y=4,xy=2,∴x2+y2+3xy,=(x+y)2+xy,=42+2,=18.

已知X2+Y2+4=2X+XY+2Y,则X2Y的值是多少?

由题意得(x-2)平方+(y-2)平方+(x-y)平方=0,故x=y=2,故x平方y=8

2x2+xy-3y2+x+4y-1因式分解

2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问:

当x=2,y=1 求{(x2+y2)/(x2-2xy+y2)+(2)/xy÷[(1/x)-(1/y)]}÷(x+y)的值

x^2+y^2=5;x^2-2xy+y^2=1;2/xy/[(1/x)-(1/y)]=-2;最后得到值为1

已知x(x-1)+(y-x2)=6,则代数式x2+y2/2-xy的值是

您好:x(x-1)+(y-x2)=6x²-x+y-x²=6y-x=6x2+y2/2-xy=1/2(x²+y²-2xy)=1/2(x-y)²=1/2*6

先化简再求值2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)],其中x=-1,y=-2.

原式=2x2-2xy-6x2+9xy-2x2+2x2-xy+y2=-2x2+5xy+2y2,当x=-1,y=-2时,原式=-2+10+8=16.

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²