（1）原式=[1/5(x²-5xy)+(5y²-xy)]÷(5y-x)
=[1/5·x(x-5y)+y(5y-x)]÷(5y-x)
=-x/5+y
(3)原式=（2a-b)²-(2a-b)
=(2a-b)(2a-b-1)
(4)n(x-m)²(x+y)+m(m-x)(x+y)²
=(x-m)(x+y)[n(x-m)-m(x+y)]
=(x-m)(x+y)(nx-mn-mx-my)
(3y-4)²-(y-1)²=0
(3y-4)²=(y-1)²
3y-4=y-1或3y-4=1-y
y=3/2或y=5/4
希望能解决您的问题.