问题
解答题
计算: (1)
(2)
(3)已知:x2+x-1=0,求x(1-
|
答案
(1)原式=
-4 a-2
+3 a-2 1 (a-2)(a+1)
=
+1 a-2 1 (a-2)(a+1)
=
+a+1 (a-2)(a+1) 1 (a-2)(a+1)
=
;a+2 (a-2)(a+1)
(2)原式=
•(a-4 (a+3)(a-3)
+a2-8a+16 (a-4)2
)•(a-3)10a-19 (a-4)2
=
•a-4 (a+3)(a-3)
•(a-3)a2+2a-3 (a-4)2
=
;a-1 a-4
(3)原式=
•x(x+1) x-1
-1 x+1 x(x+1) x-1
=
.-x2 x-1
∵x2+x-1=0,
∴x-1=-x2,原式=
=1-x2 -x2