sos!分式的加减(1) 1/m2-4 + 1/m+2 + 1/2-m(2) x2-y2/x+y ÷ (2 - x2+y2/xy)(3) ( x/x-2 - x/x+2 ) ÷ 4x/2-x (4) a2+a/a-1 ÷ (a- a/a-1 )(5) x/x2-1 .x2+2x+1/(x-2)(x+1) ÷ 2x/x-1(6) 1/m-3 - 1/m2-9 ÷ m-1/6+2m(7)x-3/x2-1 ÷ x2-2x-3/x2+2x+1 + 1/x-1

sos!分式的加减(1) 1/m2-4 + 1/m+2 + 1/2-m(2) x2-y2/x+y ÷ (2 - x2+y2/xy)(3) ( x/x-2 - x/x+2 ) ÷ 4x/2-x (4) a2+a/a-1 ÷ (a- a/a-1 )(5) x/x2-1 .x2+2x+1/(x-2)(x+1) ÷ 2x/x-1(6) 1/m-3 - 1/m2-9 ÷ m-1/6+2m(7)x-3/x2-1 ÷ x2-2x-3/x2+2x+1 + 1/x-1
sos!分式的加减
(1) 1/m2-4 + 1/m+2 + 1/2-m
(2) x2-y2/x+y ÷ (2 - x2+y2/xy)
(3) ( x/x-2 - x/x+2 ) ÷ 4x/2-x
(4) a2+a/a-1 ÷ (a- a/a-1 )
(5) x/x2-1 .x2+2x+1/(x-2)(x+1) ÷ 2x/x-1
(6) 1/m-3 - 1/m2-9 ÷ m-1/6+2m
(7)x-3/x2-1 ÷ x2-2x-3/x2+2x+1 + 1/x-1

sos!分式的加减(1) 1/m2-4 + 1/m+2 + 1/2-m(2) x2-y2/x+y ÷ (2 - x2+y2/xy)(3) ( x/x-2 - x/x+2 ) ÷ 4x/2-x (4) a2+a/a-1 ÷ (a- a/a-1 )(5) x/x2-1 .x2+2x+1/(x-2)(x+1) ÷ 2x/x-1(6) 1/m-3 - 1/m2-9 ÷ m-1/6+2m(7)x-3/x2-1 ÷ x2-2x-3/x2+2x+1 + 1/x-1
解(1)
原式=1/(m²-4)+1/(m+2)+1/(2-m)
=1/[(m+2)(m-2)]+1/(m+2)-1/(m-2)
=1/[(m+2)(m-2)]+(m-2)/[(m+2)(m-2)]-(m+2)/[(m+2)(m-2)]
=[1+(m-2)-(m+2)]/[(m+2)(m-2)]
=(1+m-2-m-2)/[(m+2)(m-2)]
=-3/[(m+2)(m-2)]
解(2)
原式=[(x²-y²)/(x+y)]÷[2-(x²+y²)/(xy)]
=[(x+y(x-y)/(x+y)]÷[(2xy/xy)-(x²+y²)/(xy)]
=(x-y)÷[-(x²-2xy+y²)/(xy)]
=(x-y)×[-xy/(x²-2xy+y²)]
=(x-y)×[-xy/(x-y)²]
= -xy/(x-y)
解(3)
原式=[x/(x-2)-x/(x+2)]÷[4x/(2-x)]
={x(x+2)]/[(x+2)(x-2)]-x(x-2)/[(x+2)(x-2)]}×[-(x-2)/(4x)]
={[x(x+2)-x(x-2)]/[(x+2)(x-2)}×[-(x-2)/(4x)]
={(x²+2x-x²+2x)/[(x+2)(x-2)]}×[-(x-2)/(4x)]
={(4x)/[(x+2)(x-2)]}×[-(x-2)/(4x)]
= -1/(x+2)
解(4)
原式=[(a²+a)/(a-1)]÷[a-a/(a-1)]
=[a(a+1)/(a-1)]÷[a(a-1)/(a-1)-a/(a-1)]
=[a(a+1)/(a-1)]÷{[a(a-1)-a]/(a-1)]
=[a(a+1)/(a-1)]÷[(a²-a-a)/(a-1)]
=[a(a+1)/(a-1)]×[(a-1)/(a²-2a)]
=[a(a+1)/(a-1)]×(a-1)/[a(a-2)]
=(a+1)/(a-2)
解(5)
原式=[x/(x²-1)]×[(x²+2x+1)/(x-2)(x+1)]÷[2x/(x-1)]
=[x/(x+1)(x-1)]×[(x+1)²/(x-2)(x+1)]×[(x-1)/(2x)]
=1/[2(x-2)]
解(6)
原式=1/(m-3)-[1/(m²-9)]÷[(m-1)/(6+2m)]
=1/(m-3)-[1/(m+3)(m-3)]×[2(m+3)/(m-1)]
=1/(m-3)-[2/(m-3)(m-1)]
=(m-1)/(m-3)(m-1)-[2/(m-3)(m-1)]
=[(m-1)-2]/[(m-3)(m-1)]
=(m-3)/[(m-3)(m-1)]
=1/(m-1)
解(7)
原式=[(x-3)/(x²-1)]÷[(x²-2x-3)/(x²+2x+1)]+1/(x-1)
=[(x-3)/(x+1)(x-1)]÷[(x-3)(x+1)/(x+1)²]+1/(x-1)
=[(x-3)/(x+1)(x-1)]×[(x+1)/(x-3)]+1/(x-1)
=1/(x-1)+1/(x-1)
=2/(x-1)