先化简,再求代数式(x-y分之一)-搜答案-智慧作业帮

将x^2+y^2+5/4=2x+y化为将x^2-2x+1+y^2-y+1/4=0即(x-1)^2+(y-1/2)^2=0上面的等式成立,则有x-1=0且y-1/2=0即x=1,y=1/2所以xy/(x

x(x-1)-(x^2-y)+2=0,开展得x^2-x-x^2+y+2=0y-x+2=0y-x=-22分之x^2+y^2再减去xy=1/2(x-y)^2=1/2*4=2再问：2分之x^2+y^2再减去

能把问题说得明白点么是不是二分之x的方加y的方这是一个整体?然后再减xy?

由已知条件可得,3x-3y=2.得,3x=3y+2代数式4x+5y+2-x+2y=3x+7y+2,将3x=3y+2代入得,10y+4即代数式的值为10y+4没办法,只能是这样表示了哦!

x=1/3y=1/2x-(x+y）+（x+2y)-(x+3y)+...-(x+13y)=6y-13y=-7y=-7/2=-3.5

3x²+xy-2y²=0(x+y)(3x-2y)=0x=-y,或3x=2yy/x=-1,或3/2x/y-y/x-(x²+y²)/(xy)=x/y-y/x-x/y

2x-y分之x+y=2,则2x-y=2(x+y)4x-2y分之x+y-4x+4y分之2x-y=2(2x-y)分之x+y-4(x+y)分之2x-y=4(x+y)分之x+y-4(x+y)分之2(x+y)=

x+y分之4x-2y-2x-y分之4x+4y=2(2x-y)/(x+y)-4(x+y)/(2x-y)=2*2-4*(1/2)=4-2=2

解原式=2(x-y)/(x+y)-(x+y)/3(x-y)=6(x-y)(x-y)/[3(x+y)(x-y)]-(x+y)x+y)/[3(x-y)(x+y)]=5(x-y)(x-y)/[(x+y)(x

因为X+2/X-Y=2所以X=2Y+2将X=2Y+2代入X+Y/3（X-Y）-（X-Y）/2（X+Y）=（3Y+2）/3（2Y+2-Y）-（2Y+2-Y）/2（3Y+2）接下来自己算

(1-x+2分之3)除以x+2分之1-x的平方=(x+2-3)/(x+2)÷(x-1)²/(x+2)=(x-1)/(x-1)²=1/(x-1)

x^2+y^2+5/4=2x+y(x-1)^2+(y-1/2)^2=0x=1,y=1/2xy/(x+y)=1*(1/2)/(1+1/2)=1/3

x=1/5y=1/7(1/x+1/y)/(1/x-1/y)=(5+7)/(5-7)=12/(-2)=6

x+y=10x-y=8/7原式=-7³/8*(x+y)³*(x-y)*4(x+y)²(x-u)²=-343/2*(x+y)^5*(x-y)³=-343

x²-x-x²+y=2;∴y-x=2;原式=（3x²+3y²）/2-3xy=(3x²+3y²-6xy)/2=3(x-y)²/2=(

将x^2+y^2+5/4=2x+y化为将x^2-2x+1+y^2-y+1/4=0即(x-1)^2+(y-1/2)^2=0上面的等式成立,则有x-1=0且y-1/2=0即x=1,y=1/2所以xy/(x

再答：不明白的地方可以追问，望采纳，谢谢

因为x＋y/2=3x＋4y/5=1,所以x＋y=2,3x＋4y=5所以解方程组得x=3,带入x＋5y＋6/3x－y＋1,得4/11

(3x+y-6)/(4x-y-8)=[3x+y-3(x-2y)]/[4x-y-4(x-2y)]=5y/7y=5/57

负11