题目 #21392

_{x−1},由x =\(2024^{0}\)+\(2^{−1}\)= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}\(\times _{x−2}= 3 3\)x2−2x x−2 x+1\(x \frac{x(x−2)}{x−3} 1 3 2 2 2\)原式=\(x^{2}\)−6−3(x−2)_{2} ⋅ (x+2)(x−2) = x2−\(3x _{2}\)⋅ (x+2)(x−2)=_{(x−2)2} ⋅ (x+2)(x−2) = x(x+2)_{x−2}=\(x_{x−2}\), ∵要使原分式有意义, ∴2+2x x的值不能取2,3, 所以x可取的值为1, 当x = 1时,\(\frac{x(x−3)}{1}\)原式= 1+2 （x−2） （x−2） x−3 x−\(3 _{1−2}\)=−3. 原\(\frac{6.6_{2}_{7}}{9.6}\)式= 1 − _{a}⋅ _{(a−3)(a+3)}= 1 − _{a+3}= _{a+3}, 当a = 4时,原式= _{4+3}= _{7}. 原式= 4 −\(a^{2}\)−\(2a^{2}\)−6a +\(3a^{2}\)= 4 −6a. 当a =− _{3}时, 原式= 6. a−3 a(a+1) a+1 3 −\(\frac{2}{b}2\)时,原式= 3 2 2