计算.(1)a+b+2b
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计算.
(1)a+b+
(1)a+b+
2b
(1)a+b+
2b2 a−b = (a+b)(a−b) a−b+ 2b2 a−b = a2+b2 a−b; (2)(x+1− 3 x−1)÷ x+2 2x−2 =[ (x+1)(x−1) x−1- 3 x−1]× 2(x−1) x+2 = (x+2)(x−2) x−1× 2(x−1) x+2 =2x-4; (3)(2+ 1 a−1− 1 a+1)÷(a− a 1−a2), =[ 2(a+1)(a−1) (a+1)(a−1)+ (a+1) (a−1)(a+1)- a−1 (a+1)(a−1)]÷[ a(a+1)(a−1) (a+1)(a−1)+ a (a+1)(a−1)] = 2a2 (a+1)(a−1)× (a+1)(a−1) a3 = 2 a, 把a=2代入原式得:原式= 2 a=1.
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