1.(3x-y-1)=

y(x+y)+(x-y)²-（x+y）（x-y）,其中x=-1/3,y=3 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 [(1)/(x-y)+(1)/(x+y)]÷（x的平方y)/(x的平方-y的平方),其中x=根号3+1,y=根号3-1. {3(x+y)+3(y+x)=1,3(x+y)+4(y-x)=-1 (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 y'=(x-y+1)/(x+y-3)通解 (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) 2(x+y)-3(x-y)=1 6(x+y)+(x-y)=51 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 变量x,y满足约束条件,x+y>=3,x-y>=-1,2x-y 二元一次方程 ：2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 (x-y)^2+(x+y)(x-y) 其中 X =3 Y=-1 x+y+2(-x-y+1)=3(1-y-x)-4-(y+x-1),x+y=? {(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=? 已知方程组{2x+y=6,x-3y=1,不解方程组,求7y(x-3y)²-(3y-x)²的值.已知方程组{2x+y=6,{x-3y=1.不解方程组,求7y(x-3y)²-(3y-x)²的值.已知方程组{2x+y=6，{x-3y=1.不解方程组，求7y(x-3y)²-(3y-x)³ y=(x-3)/(x-2)yx-2y=x-3(y-1)x=2y-3x=(2y-3)/(y-1)∴y≠1 问：为什么在x=(2y-3)/(y-1)这一步才y≠1 ,而不在(y-1)x=2y-3时讨论y是否等于1..依据是什么? 已知：2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x) 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1