Question

求解以下一道二次根式题

Answer 1

为使√(x-a)与√(a-y)有意义, 有y ≤ a ≤ x.

而a, x, y互不相等, 故y < a < x.

由x-a > 0, 为使√(a(x-a))有意义, 有a ≥ 0.

又y-a < 0, 为使√(a(y-a))有意义, 有a ≤ 0.

于是只有a = 0, 代回等式得0 = √x-√(-y), 可知x = -y.

代入得(3x²+xy-y²)/(x²-xy+y²)

= (3(-y)²+(-y)y-y²)/((-y)²-(-y)y+y²)

= y²/(3y²)

= 1/3.