问题
解答题
计算题: (1)
(2)
(3)|-3|+(-1)2013×(π-3)-(-
|
答案
(1)原式=5x+3y-2x x2-y2
=3(x+y) (x+y)(x-y)
=
;3 x-y
(2)原式=
+x-1 (x+1)(x-1) 2 (x+1)(x-1)
=x+1 (x+1)(x-1)
=
;1 x-1
(3)原式=3-1×(π-3)+8
=3-π+3+8
=14-π.
计算题: (1)
(2)
(3)|-3|+(-1)2013×(π-3)-(-
|
(1)原式=5x+3y-2x x2-y2
=3(x+y) (x+y)(x-y)
=
;3 x-y
(2)原式=
+x-1 (x+1)(x-1) 2 (x+1)(x-1)
=x+1 (x+1)(x-1)
=
;1 x-1
(3)原式=3-1×(π-3)+8
=3-π+3+8
=14-π.