x=1,y=1,x=2,y=3,x=3,y=6,x=4,y=10是怎么推出y=x(x+1)/2的如题可是就算有大致图像,他也可能不一定是二次函数啊

若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 二元一次方程 ：2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 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 (x-y)^2+(x+y)(x-y) 其中 X =3 Y=-1 (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) {(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=? x+y+2(-x-y+1)=3(1-y-x)-4-(y+x-1),x+y=? [3x^2-(x+y)(y-x)][2(-x-y)(y-x)+2x^2+3y^2],其中x=-2分之1,y=1.5 已知：2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x) y(x+y)+(x-y)²-（x+y）（x-y）,其中x=-1/3,y=3 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 先化简 再求值 (2x-y)(y+x)-(x-2y)(2y+x)-(-3y+x)^2其(√x+1)+y^2+4=-4y 已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙 已知x-y=1,求[(x+2y)²+(2x+y)(x-4y)-3(x+y)(x-y)]÷y的值 函数,y=3x/(x^2+x+1) ,x