若x²+y²=12,x y=4那么(x-y)²=?

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若绝对值x-1,+y的平方+4y+4=0,求xy的值

|x-1|+y^2+4y+4=|x-1|+(y+2)^2=0因为|x-1|>=0,(y+2)^2>=0则:x-1=0y+2=0x=1y=-2xy=-2

已知x²+y²=7,xy=-2.那么5x²-3xy-4y²-11xy-7x

5x²-3xy-4y²-11xy-7x²+2y²=5x²-7x²+2y²-4y²-3xy-11xy=-2x²-

(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=

(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=-9xy+(x-2y)(x-2y-x-2y)=-9xy+(x-2y)(-4y)=-9xy-4xy+8y²=8y

若正实数x ,y满足2x+y+6=xy.则xy的最小值.

2x+y+6≥6+2√2xyxy≥6+2√2xy(√xy-√2)^2≥8√xy-√2≥2√2或√xy-√2≤-2√2(不可能)所以xy最小值是(3√2)^2=18-------------------

若2x+y=0,则x²+xy+y²/2xy-x²的值( )

即y=-2x所以原式=(x²-2x²+4x²)/(-4x²-x²)=3x²/(-5x²)=-3/5

若x>1,y>0且满足xy=xy,xy=x

由题设可知y=xy-1,∴x=yx3y=x4y-1,∴4y-1=1,故y=12,∴12x=x,解得x=4,于是x+y=4+12=92.故答案为:92.

已知:x+y=1,xy=−12

x(x+y)(x-y)-x(x+y)2=x(x+y)(x-y-x-y),=x(x+y)(-2y),=-2xy(x+y),当x+y=1,xy=-12时,原式=-2×(-12)×1=1.

已知X方+4XY+Y方=0 求X+Y分之X-Y

像这种题目也可以用特殊值去做,比如令x=1,代入方程求出y,再代入所求的式子就可以得出答案.

xy=-2,x+y=3,求代数式(4xy+12y)+[7x-(3xy+4y-x)]

(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

若x-y=4,xy=1,求(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

已知2x²+xy=10,3y²+2xy=6,求4x²+8xy+9y²的值

4x²+8xy+9y²=2(2x²+xy)+3(3y²+2xy)=38.

已知x+y=0,x+13y=1,求x²+12xy+13y²的值.

解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+

若x+y=5,xy=6,求x的平方y+xy的2次方的值

求x的平方y+xy的2次方=xy(x+y)=30

已知x大于0,y大于0,x+y=4,求xy的最大值

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,

数学已知A=5x的平方y-3xy平方+4xy,B=7xy平方-2xy+x平方y,若A+B+2C=0化简C-A

A=5x^2y-3xy^2+4xyB=7xy^2-2xy+x^2y2C=-A-B2C-2A=-3A-B=-15x^2y+9xy^2-12xy-7xy^2+2xy-x^2y=-16x^2y+2xy^2-

已知x+y=3,xy=-2,求(3x-2y+8xy)-(x-4y-5xy)的值.

(3x-2y+8xy)-(x-4y-5xy)=3x-2y+8xy-x+4y+5xy=(3x-x)+(-2y+4y)+(8xy+5xy)=2x+2y+13xy=2(x+y)+13xy=2×3+13×(-

若x+y=5,xy=3,则x^2·y+2x^2·y^2+xy^2=

x^2·y+2x^2·y^2+xy^2=xy×(x+2xy+y)=3×(5+6)=33谢谢~

x^2+y^2+4x+6Y+13=0中xy是多少?

x^2+y^2+4x+6y+13=0x^2+y^2+4x+6y+4+9=0x^2+4x+4+y^2+6y+9=0(x+2)^2+(y+3)^2=0(∵(x+2)^2>=0,(y+3)^2>=0)=>(

若有理数数x,y满足xy≠0,则m=x|x|+|y|y

∵有理数x,y满足xy≠0,∴x|x|=±1,|y|y=±1,∴m=x|x|+|y|y的最大值是m=1+1=2.故答案为:2.