一道一元二次方程的题,急啊,赶快,谢谢!已知m是一元二次方程x^2+3x-1=0的实数根,求代数式(m-3)/(3m^2-6m)÷[m+2-5/(m-2)]的值.（如果看不懂的话就看图）赶快啊,谢谢

Question

一道一元二次方程的题,急啊,赶快,谢谢!
已知m是一元二次方程x^2+3x-1=0的实数根,求代数式
(m-3)/(3m^2-6m)÷[m+2-5/(m-2)]的值.
（如果看不懂的话就看图）
赶快啊,谢谢

Answer 1

m是一元二次方程x^2+3x-1=0的实数根 则m²+3m-1=0 即m（m+3）=1
(m-3)/(3m^2-6m)÷[m+2-5/(m-2)]=(m-3)/3m(m-2)÷[（m²-4）/(m-2)-5/(m-2)]
=(m-3)/3m(m-2)÷[（m²-9）/(m-2)]
=(m-3)/3m(m-2)÷[（m+3）（m-3）/(m-2)]
=(m-3)/3m(m-2)*[(m-2)/（m+3）（m-3）]
=1/3m(m+3)
=1/3