When X=8, obviously this equation does not hold.

When x is not 8

Ensure that X> when y is 0, it is also greater than 0.

Obviously, let X> VIII

X+Y=X+2X/(X-8)

Let this value be k.

Then x(x-8)+2X=K(X-8)

X^2-(6+k)x+8K=0

In other words, this equation has roots greater than 8.

δ=k^2+ 12k+36-32k=k^2-20k+36≥0

The solution is k≥ 18.

(If K is less than or equal to 2, because X is already greater than 8, plus a Y is greater than 0, obviously K must be greater than 8. )

When k≥ 18, the symmetry axis is (18+6)/2 =12 > eight

So the equation can also guarantee that there must be a root greater than 8.

Therefore, the minimum value of X+Y is 18.

At this time, X= 12 and Y=6.

I suggest LZ use my method in the exam. Although the method of building 1 is ingenious and difficult to think about, my method is gradual and will be achieved. In case of mistakes, there must be steps. In fact, the building 1 is also wrong. If you add up his last x and y, it must not be 18.