从圆x^2+y^2-2x-2y+1=0外一点(3,2)向这个圆做两条切线,则两切线夹角的余弦值

x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 设x大于1,y大于0,x^y+x^-y=2根号二,x^y-x^-y等于? 从M={(x,y)||x-2|+|y-2| 2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2 （x-y)(x+y)-(x+y)^2+2y(y-x),其中x=1,y=3. mathematica软件,已知y[x],Y=f1(x),X(x)=f2(x),如何plot Y[X]?y[x_] := x + 9.81/2*x*xk=1X[x] = x - k*y'[x]/(1 + (y'[x])^2)^0.5Y[x] = y[x] + k/(1 + (y'[x])^2)^0.5Plot[{y[x],Y[X],},{x,0,2}] 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 [(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y 已知x,y满足约束条件：x-y+1>=0,x+y-2>=0,x 若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值. 已知x>0,Y>0,如何证x^2/y+y^2/x>=x+y x+2y=2x+y+1=7x-y 求：2x-y? (X,Y) f(x,y)={12y^2,0 二元一次方程 ：2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 (3x-y)^2+(3x+y)(3x-y),x=1,y=-2 已知2x-y=0,求分式【1+（3*y*y*y/x*x*x-y*y*y）】/【1+（2y/x-y）】的值注：x*x*x-y*y*y=（x-y）*（x*x+xy+y*y） x*x+2x+y*y-4y+5=0 x= y=