(1)设原式=(ax+2)²
则a²=2m-1,(m+1)/2=a
所以(m+1)²/4=2m-1
m²+2m+1=8m-4
m²-6m+5=0
(m-1)(m-5)=0
所以m=1或5
(2)(x-y)²+y-x=1
(x-y)²-(x-y)-1=0
b²-4ac=5
x-y=(1加减(根号5))/2
(1)设原式=(ax+2)²
则a²=2m-1,(m+1)/2=a
所以(m+1)²/4=2m-1
m²+2m+1=8m-4
m²-6m+5=0
(m-1)(m-5)=0
所以m=1或5
(2)(x-y)²+y-x=1
(x-y)²-(x-y)-1=0
b²-4ac=5
x-y=(1加减(根号5))/2