如果方程组x 2y=k,2x y=1
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(x+y)(xy)=x^2y+xy^2=-8原式=-7
解出x=(7k-4)/5y=(-8k+11)/5由x>yk>1
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.
[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.
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
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.
解-x²y-xy²=-xy(x+y)=-2×5=-10
∵x+y=0,xy=-7,∴①x2y+xy2=xy(x+y)=-7×0=0;②x2+y2=(x+y)2-2xy=14.
∵2x+y=4,xy=3,∴2x2y+xy2=xy(2x+y)=3×4=12.故答案为:12
原式=y(x2+2x+1)=y(x+1)2,故答案为:y(x+1)2.
8x2y-8xy+2y,=2y(4x2-4x+1),=2y(2x-1)2.
4x-3y=7,[(k+1)/2]x-y=k-3.8x-6y=14,3(k+1)x-6y=6k-18.二式减一式等于:(3k-5)x=6k-32,x=(6k-32)/(3k-5)有得:y=(3k-27
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
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
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.
原式=-xy(x-y),当x-y=3,xy=-2时,则原式=-3×(-2)=6.故答案为:6.
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
x2y-2xy-y=y(x2-2x-1)=y(x2-2x+1-2)=y[(x-1)2-(2)2]=y(x-1+2)(x-1-2),故答案为:y(x-1+2)(x-1-2).
解;∵x+y=0,xy=-7∴x2y+xy2=xy(x+y)=-7×0=0x2+y2=(x+y)2-2xy=02-2×(-7)=0+14=14.