已知m是二元一次方程x²+3x-1=0的实数根,求代数式[(m-3)/(3m²-6m)]/[(m+2)-5/(m-2)]

如题所述

∵m是方程x²+3x-1=0的实数根
∴m²ï¼‹3m-1=0
∴m²ï¼‹3m=1
[(m-3)/(3m²-6m)]/[(m+2)-5/(m-2)]
=[(m-3)/[3m(m-2)]/﹙m²ï¼9﹚/(m-2)]
=[(m-3)/[3m(m-2)]·(m-2)/[﹙m+3﹚﹙m-3﹚]
=1/[3m﹙m+3﹚]
=1/[3﹙m²ï¼‹3m﹚]
=1/3
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第1个回答  2012-12-04
代数式[(m-3)/(3m²-6m)]/[(m+2)-5/(m-2)]=1/[3m(m+3)],方程的根为(-3±√13)/2,分别将两根代入代数式,得1/3
第2个回答  2012-12-04
先将分式化简成最简形式,1/[3(m²+3m)]
将m带入方程得m²+3m=1,
带入分数式的1/3