y=x^2-2x-5,x∈[-1,2],求值域有没有简单易懂的方法?刚学还不太明白

x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 {6(x-y)-7(x+y)=21 {2(x-y)-5(x+y)=-1 (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) 若2x-3y+4=0则x（x*x-1)+x(5-x*x)-6y+7 {(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=? 分式的加减:[(x/y-y/x)除以（x+y）+x（1/y-1/x)]除以1+x/y,其中x=5,y=2 已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y) 集合A={y y=x2+2x-1,x∈R},B={x (2-x)(x+5) （2x+y-1)(2x-y+1)-(3x+y)(3x-y),其中x=5分之1, 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 函数,y=3x/(x^2+x+1) ,x 函数y=3x/(x^2+x+1) （x 函数y=2x/x²-x+1（x y=2x+1,x∈Z且|x| 求y=(5x-1)/(4x+2) (x 已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1