已知x2 2y2 z2-2xy-2y-4z 5=0-搜答案-智慧作业帮

由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1

y=3+xyy-xy=3y(1-x)=3y=3/(1-x)1)2x+3xy-2y/x-2xy-y=2x+xy-2y/x-y=2x+3x/(1-x)-6/x(1-x)-3/(1-x)=2x-3/(1-x

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

xy^2*(-x^2y^2)*（1/2xy^3)^2=xy^2*(-x^2y^2)*（1/4x²y^6)=xy^2*(x^4y^8)*(-1/4)=(-1/4)*xy²*(xy&#

因为x/y=2所以x=2y是[x^2-xy+3y^2]/[x^2+xy+6y^2]吧=(4y^2-2y^2+3y^2)/(4y^2+2y^2+6y^2)=5y^2/12y^2=5/12

-xy（x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

x=-1y=-2或x=-2y=-1-2（xy-3x）-2（2y-xy）=-2（xy-3x+2y-xy）=-2（2y-3x）等于2或-8

（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

把已知两式相加结果为13

答案是3/2你是不是把分母打错了教你个方法因为上下是齐次的直接令x=3y=2带入就行

xx+2xy-yy=-3两式相加即可

2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3（x+y）+2xy=3*（-2）+2*3=0

z=3x+y=13(x+2y)/6+5(x-4y)/6当x=5,y=2时取到,z最大值17

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

3xy=2x+3y+5(3y-2)(x-1)=7所以3y-2=7x-1=1得y=3x=2xy=6或3y-2=1x-1=7得y=1x=6xy=6所以xy=6

3xy-3xy-xy+2yx

3xy-3xy-xy+2yx=-xy+2xy=xy

x^2-xy=14,(1)xy-y^2=-11,(2)(1)-(2)得：x^2-2xy+y^2=14-(-11)=25

原式=-xy²（x²y^4-xy²-1）∵xy²=-2原式=2（（-2)²-（-2）-1）=10