若xy满足x 2y-52y=17 4根号2

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若xy满足x 2y-52y=17 4根号2
已知x+y=-5,xy=7,求x2y+xy2-x-y的值.

x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.

已知x和y是正整数,并且满足条件xy+x+y=71,x2y+xy2=880,求x2+y2的值.

∵xy+x+y=71,x2y+xy2=880,∴xy(x+y)=880,xy+(x+y)=71,∴x+y、xy可以看做一元二次方程t2-71t+880=0的两个解,解得t=55或16,∴x+y=55、

若x+y=2,xy=-4,求x2y+xy2+1的值

(x+y)(xy)=x^2y+xy^2=-8原式=-7

已知x、y均为实数,且满足xy+x+y=17,x2y+xy2=66,则x4+x3y+x2y2+xy3+y4=______

x2y+xy2=xy(x+y)=66,设xy=m,x+y=n,由xy+x+y=17,得到m+n=17,由xy(x+y)=66,得到mn=66,∴m=6,n=11或m=11,n=6(舍去),∴xy=m=

若实数x,y满足xy>0且x2y=2,则xy+x2的最小值是(  )

xy+x2=xy2+xy2+x2≥33x4y24=3当且仅当xy2=x2时成立所以xy+x2的最小值为3故选A.

已知实数x、y满足x+y+xy=9,x2y+xy2=20,求x2+y2的值.

x+y+xy=9x+y=9-xyx^2y+xy^2=20xy(x+y)=20xy(9-xy)=20xy^2-9xy+20=0(xy-4)(xy-5)=0xy=4或xy=5x+y=5或x+y=4x^2+

已知x、y均为实数,且满足xy+x+y=17,x2y+xy2=66,求x2+y2

由已知:xy+x+y=17,xy(x+y)=66,可知xy和x+y是方程t2-17t+66=0的两个实数根,得:t1=6,t2=11.即xy=6,x+y=11,或xy=11,x+y=6.x2+y2=(

先化简,后求值.[2x(x2y-xy2)+xy(xy-x2)]÷x2y,其中x=2013,y=2012.

[2x(x2y-xy2)+xy(xy-x2)]÷x2y=[2x3y-2x2y2+x2y2-x3y]÷x2y=x-y,把x=2013,y=2012代入上式得:原式=x-y=2013-2012=1.

2(x2y+xy)-3(x2y+xy)-4x2y其中x=-2,y=12

原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.

先化简后求值:4x2y-[6xy-3(4xy-2)-x2y]+1,其中x=2,y=-12

原式=4x2y-6xy+3(4xy-2)+x2y+1=5x2y+6xy-5当x=2,y=-12时,原式=5×4×(-12)+6×2×(-12)-5=-21.

若实数x,y满足xy+x+y+7=0,3x+3y=9+2xy,则x2y+xy2=______.

∵xy+x+y+7=0               

已知x+y=6,xy=-3,则x2y+xy2=

那个2是平方吧?可以用^代替原式=x^y+xy^=xy(x+y)=-3*6=-18

分解因式:x2y+2xy+y=______.

原式=y(x2+2x+1)=y(x+1)2,故答案为:y(x+1)2.

已知x、y均为实数,且满足xy+x+y=17,x2y+xy2=66,求x4+x3y+x2y2+xy3+y4的值.

方程ax^2+bx+c=0,判断这个方程有没有实数根,有几个实数根,就要用ΔΔ=b^2-4ac若Δ<0,则方程没有实数根Δ=0,则方程有两个相等实数根,也即只有一个实数根Δ>0,则方程有两个不相等的实

若x+y=5,xy=6,则x2y+xy2的值为______.

∵x+y=5,xy=6,∴x2y+xy2=xy(x+y)=5×6=30.故答案为:30.

若M,N为单项式且1/2x2y+M=1/2xy(N+2y),求M,N

1/2x²y+M=1/2xy(N+2y)=1/2xyN+xy²所以N=xM=xy²

数学竞赛题:若实数x,y满足方程组xy+x+y+7=0,3x+3y=9+2xy,则x2y+xy2=?

x2y+xy2=xy*(x+y)因为x+y=-(7+xy)又x+y=(9+2xy)\3所以(9+2xy)\3=-(7+xy)3+2xy\3=-7-xy5xy\3=-10解得xy=-6所以x+y=-(7

化简求值:2(x2y+xy)-3(x2y-xy)-4x2y,其中x=-1,y=1.

原式=2x2y+2xy-3x2y+3xy-4x2y=-5x2y+5xy,当x=-1,y=1时,原式=-5×(-1)2×1+5×(-1)×1=-5-5=-10.

若x-y=3,xy=-2,则xy2-x2y的值是______.

原式=-xy(x-y),当x-y=3,xy=-2时,则原式=-3×(-2)=6.故答案为:6.

已知X2+4y2-4x+4y+5=0 求x-y的值 已知xy=4满足x2y-xy2-x+y=56,求x2+y2的值

x²+4y²-4x+4y+5=0(x-2)²+(2y+1)²=0x-2=0x=22y+1=0y=-1/2x-y=2+1/2=5/2x²y-xy