X Y=6和x 2y=9的二元一次方程
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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.
(x+y)(xy)=x^2y+xy^2=-8原式=-7
x+y=5k(1)x-y=9k(2)(1)+(2)2x=14kx=7k(1)-(2)2y=-4ky=-2k把x=7k,y=-2k代入2x+3y=6得14k-6k=68k=6k=3/4
x+y=15kx-y=9k那么,2x=24k,2y=6k,y=3k,2x+3y=6就是2x+2y+y=624k+6k+3k=633k=611k=2k=2/11既然是选择题,你就自己一个一个选项试一试,
xy=1就是y=1/x当然是不是二元一次方程式了
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+
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=3是二元二次方程式
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.
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
原式=y(x2+2x+1)=y(x+1)2,故答案为:y(x+1)2.
再问:2y=3x是怎么来的再答:比例的性质,内项积等于外项积
(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
∵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
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