已知关于xy的不等式组x 2y=1 m,2x y=2
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x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
∵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、
(x+y)(xy)=x^2y+xy^2=-8原式=-7
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+
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
①x2y+xy2=xy(x+y)=1×3=3;②x2+y2=(x+y)2-2xy=32-2×1=7.
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
∵x+y=6,xy=4,∴x2y+xy2=xy(x+y)=4×6=24.故答案为:24.
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
∵xy<0,由二次根式的有意义,得y>0,∴x<0,∴原式=x2y=-xy.
那个2是平方吧?可以用^代替原式=x^y+xy^=xy(x+y)=-3*6=-18
解-x²y-xy²=-xy(x+y)=-2×5=-10
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
若是209,则xy=8,x+y=15,算出x,y就不是整数了,与题意不符.若是34,x,y为3,5,符合题意.
(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)/(2y)=X^2-y^2-y^2+X^2+y^2-2xy-3x^2+xy=-x^2-y^2-xy=-(x^2+y^2+xy-3
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
∵x+y=5,xy=6,∴x2y+xy2=xy(x+y)=5×6=30.故答案为:30.
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
由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6
∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a