y=-x^2-2x+3(-5≤x≤-2)y=2x^2-4x+1(-1≤x≤2)y=(1-x^2)/(1+x^2)我想知道解题的过程或思路

y=-3x,-2≤x 已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y) y=-x²-3x+5 的值域 y=2x²+3x-3,x≤【0. 3^x+2x=y,x x+y/2+x-y/6=5 2(x+y)-3(x-y)=-2 3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2 x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) 4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11 3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27 6(x-y)-2(x+y)=14 3(x-y)+(x+y)=5 先化简再求值(x-y)(x-2y)+(x-2y)(x-3y)-2(x-3y)(x-4y) x=4 y=5 先化简,再求值 [2/3x - 2/x+y（x+y/3x -x-y）]*x/x-y,其中5x+3y=0 若变量x,y满足约束条件-1≤x:x≤y;3x+2y≤5,则z=2x+y的最大值为 先化简,再求值:[3x-2/x+y(x+y/3x-x-y)]/(x-y/x),其中5x+3y=0 若变量x y满足约束条件x+y≤5 x-y≤-1 x≥0 则z=3x-2y的最小值 若2x-3y+4=0则x（x*x-1)+x(5-x*x)-6y+7 设x,y满足约束条件{x-y+5≥0,x+y≥0,x≤3}求z=(x+1)^2-y^2的最大值?