If x3 is brought into the equation: (x3-x 1)(x3-x2)=2, x3>x1x3 > X2 or x3 < x1x3 <; X2 (same number principle)

And divide by the equation (x-x 1)(x-x2)=2 to get x2-(x1+x2) x+x1x2-2 = 0.

de:x3+x4 = x 1+x2 x3x 4 = x 1x 2-2。

X3+x4=x 1+x2 minus x3 >: x1x3 > X2 can't hold on.

So: x3

Similarly, we can deduce: x4>x1x4 > X2 is x 1

So: x3