2x y)2−4(x−y)(x 2y)

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已知x2+y2-2x-4y+5=0,分式yx−xy

∵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.

x2-xy-2y2-x+5y-2

首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y-2可分解成(X-2Y+A)(X+Y+B)则展开有X的一次项=A+B=-1Y的一次项有A-2B=5连列成方程

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

若X2+Y+2x-4根号y+5=0,求根号xy

x^2+2x+1+y-4根号y+4=0(x+1)^2+(根号y-2)^2=0x+1=0,根号y-2=0x=-1,y=4题目好像打错了

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

已知代数式A=2x2+3xy+2y-1,B=x2−xy+x−12

(1)A-2B=2x2+3xy+2y-1-2(x2−xy+x−12)=2x2+3xy+2y-1-2x2+2xy-2x+1=5xy+2y-2x,当x=y=-2时,A-2B=5xy+2y-2x=5×(-2

因式分解 x2+xy-2y2-x+7y-6

x^2+xy-2y^2-x+7y-6=(x+2y)(x-y)-x+7y-6x+2y-3Xx-y2十字相乘得:2(x+2y)-3(x-y)=-x+7y是一次项所以原式=(x+2y-3)(x-y+2)

x2-2xy+y2+3x-3y+2.

x2-2xy+y2+3x-3y+2=(x-y)2+3(x-y)+2=(x-y-1)(x-y-2).

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2x2-xy-3y2=0,求x−yx+y

2x2-xy-3y2=0,(2x-3y)(x+y)=0,解得:2x-3y=0或x+y=0(分母为0,舍去),解得:x=3y2,则x−yx+y=3y2−y3y2+y=y5y=15.

已知,x和y是任意实数,M是代数式x2+2xy+y2,x2-2xy+y2,x2+4x+4中的最大值,求M的最小值

1由x2+2xy+y2+x2-2xy+y2=2x2+2y2知x2+2xy+y2和x2-2xy+y2中的最大值大于等于x2由x2+x2+4x+4=2(x+1)2+2知x2和x2+4x+4中的最大值大于等

约分 xy+2y/x2次方-4

y(x+2)/(x-2)(x+2)=y/(x-2)提取yx²-4=(x-2)(x+2)

先化简,后求值:4xy-(2x2+5xy-y2)+2(x2+3xy),其中|x+2|+(y−12)

4xy-(2x2+5xy-y2)+2(x2+3xy),=4xy-2x2-5xy+y2+2x2+6xy),=5xy+y2,∵.x+2  .+(y-12)2=0∴x=-2,y=12当x

正数xy满足x2-y2=2xy,求x+y分之x-y的值

x^2-y^2=2xy,得x/y-y/x=2,即(y/x)^2+2(y/x)-1=0∴y/x=-1+√2或y/x=-1-√2(舍去,因为x,y都是正数).即(x-y)/(x+y)=√2-1

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:

2x(x-y)4-x2(x-y)2+xy(y-x)2 如何因式分解

原式=2x(x-y)4-x2(x-y)2+xy(x-y)2=x(x-y)2{2(x-y)2-x+y}=x(x-y)2{2(x-y)2-(x-y)}=x(x-y)3(2x-2y-1)

x2-xy-2y2-x+5y-2因式分解

首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y-2可分解成(X-2Y+A)(X+Y+B)则展开有X的一次项=A+B=-1Y的一次项有A-2B=5连列成方程

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

已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]

x^2+y^2-2x+4y+5=0(x-1)^2+(y+2)^2=0x=1,y=-2化解式子,再代入