Question

数学分式计算

Answer 1

∵m是方程x²+3x-1=0的根

∴m²+3m-1=0
∴m²+3m=1

原式=(m-3)/3m(m-2) ÷[(m+2)(m-2)/(m-2) -5/(m-2)]
=(m-3)/3m(m-2)÷(m²-4-5)/(m-2)
=(m-3)/3m(m-2)×(m-2)×(m+3)(m-3)
=1/3m(m+3)
=1/3(m²+3m) m²+3m=1
=1/3

Answer 2

x²+3x-1=0即x²+3x=1
(m-3)/(3m²-6m)÷[m+2-5/(m-2)]
=(m-3)/(3m²-6m)÷[(m+2)(m-2)/(m-2)-5/(m-2)]
=(m-3)/(3m²-6m)÷[(m²-4)/(m-2)-5/(m-2)]
=(m-3)/(3m²-6m)÷[(m²-4-5)/(m-2)]
=(m-3)/(3m²-6m)÷[(m²-9)/(m-2)]
=(m-3)/(3m²-6m)÷[(m+3)(m-3)/(m-2)]
=[(m-3)/(3m²-6m)]*{(m-2)/[(m+3)(m-3)]}
=[1/(3m²-6m)]*[(m-2)/(m+3)]
=(m-2)/[(3m²-6m)(m+3)]
=(m-2)/[3m(m-2)(m+3)]
=1/[3m(m+3)]
=1/[3(m²+3m)]
=1/3