圆x^2+y^2+2x-2y+1=0被直线x-y+3=0截得的劣弧长为?类似题的中心解题思想?
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若x-3y=0,且x≠0,2x+5y/3x-y的直是?
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
已知X,Y为实数,且X=根号1-3Y加根号3Y-1加2,求6Y-4X的直
2x-3y=3,4x-3y=15.求x,y的直(用消元)
设x大于1,y大于0,x^y+x^-y=2根号二,x^y-x^-y等于?
方程(x+y-1)(根号下x2+y2-4) =0的曲线形状是什么?由方程:(x+y-1)*√(x²+y²-4) =0,可得(1) x+y-1=0,且x²+y²-4 ≥ 0(2) x²+y²-4=0 ;由(1) 可知x+y=1是直线;直线穿过圆,被圆周分为3部分;直
已知实数x,y满足关系式x2+y2-6x-4y+12=01)y/x的最大直 和最小直2)x-y 的最大直 和最小直3)x^2 +y^2的最大直 和最小直
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