Solution: 1, [x(x-3)]/(x? +2) = 0 = "x (x-3) = 0 =" x = 0 or x=3.

2x/(x？ - 1)+2x/( 1-x)=2x/(x？ - 1)-2x(x+ 1)/(x？ - 1)=2x？ /( 1-x？ )

When x=0, 2x/(x? - 1)+2x/( 1-x)=0

When x=3, 2x/(x? - 1)+2x/( 1-x)=-9/4

2 、[(2x+y )/(x？ -2xy+y？ )] × (x-y)=[(2x+y )/(x-y)？ ]×(x-y)= =(2x+y)/(x-y)

Because x-3y = 0 = "x = 3y.

So [(2x+y )/(x? -2xy+y？ )]×(x-y)=(2x+y)/(x-y)=(6y+y)/(3y-y)= 7/2