问题
解答题
先化简,再求值:
|
答案
原式=
+x x+2
×x(x-2) (x+1)(x-1)
=x+1 x(x-2)
+x x+2
=1 x-1
,x2+2 (x+2)(x-1)
∵x2+x-1=0,
∴x2+x=1,x=
,-1± 5 2
∴原式=
=x2+2 x2+x-2
=-x2-2=x2+2 1-2
.-1± 5 2
先化简,再求值:
|
原式=
+x x+2
×x(x-2) (x+1)(x-1)
=x+1 x(x-2)
+x x+2
=1 x-1
,x2+2 (x+2)(x-1)
∵x2+x-1=0,
∴x2+x=1,x=
,-1± 5 2
∴原式=
=x2+2 x2+x-2
=-x2-2=x2+2 1-2
.-1± 5 2