解不等式(1)(y+1/6)-(2y-5/4)>=1+(y/2) (2)(x-3/2)+3>=x 1-3(x-1)

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-( 解方程组：2(x-y)/3-x+y/4=-1 ,6(x+y)-4(2x-y)=16 解二元一次方程组{3(x+y)-4(x-y)=4,x+y/2+x-y/6=1 解方程组3(x+y)-4(x-y)=4和(x+y)/2+(x-y)/6=1 解二元一次方程组{3(x+y)-(x-y)=4,x+y/2+x-y/6=1 解二元一次方程组：3(x+y)-4(x-y)=4,x+y/2+x-y/6+1 (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 （x+2)(x+3) (x-4)(x+1) (y+4)(y-2) (y-5)(y-3) 3x+2y/4=2x+y/5=x-y+1/6