Question

已知：a是一元二次方程X^2+3X-1=0的实数根，求代数式

图片中的第二题！

Answer 1

整理（a-3）/(3a²-6a)÷[a+2-5/(a-2)]
=（a-3）/3a（a-2)÷[(a+2)(a-2)/a-2-5/(a-2)]
=（a-3）/3a（a-2)÷[(a+2)(a-2)-5/(a-2)]
=（a-3）/3a（a-2)÷[(a²-9)/(a-2)]
=（a-3）/3a（a-2)*(a-2)/(a+3)(a-3)
=1/3a(a+3)
=1/3(a²+3a)
因为a是一元二次方程X^2+3X-1=0的实数根
所以a²+3a-1=0，
所以a²+3a=1,代人，得，
原式=1/3

Answer 2

x=a代入方程：a²+3a-1=0
a²+3a=1

[(a-3)/(3a²-6a)]÷[a+2 -5/(a-2)]
=[(a-3)/3a(a-2)]/[(a²-9)/(a-2)]
=[(a-3)/3a]/[(a+3)(a-3)]
=1/[3a(a+3)]
=1/[3(a²+3a)]
=1/(3×1)
=1/3