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九年级数学解答题一般
题目

先化简,再求值:(2a+1+a+2a2−1)÷aa+1\left(\dfrac{2}{a+1}+\dfrac{a+2}{a^{2}-1}\right)\div \dfrac{a}{a+1},其中a=2sin⁡60∘+tan⁡45∘a=2\sin 60^{\circ}+\tan 45^{\circ}.

知识点：分式的化简求值章节：未标注

答案与解析

答案

原式=[2a−2(a+1)(a−1)+a+2(a+1)(a−1)]⋅a+1a=\left[\dfrac{2a-2}{\left(a+1\right)\left(a-1\right)}+\dfrac{a+2}{\left(a+1\right)\left(a-1\right)}\right]\cdot \dfrac{a+1}{a}

=3a(a+1)(a−1)⋅a+1a=\dfrac{3a}{\left(a+1\right)\left(a-1\right)}\cdot \dfrac{a+1}{a}

=3a−1=\dfrac{3}{a-1}，

当a=2sin⁡60∘+tan⁡45∘=2×32+1=3+1a=2\sin 60^{\circ}+\tan 45^{\circ}=2\times \dfrac{\sqrt {3}}{2}+1=\sqrt {3}+1时，

原式=33+1−1=3=\dfrac{3}{\sqrt {3}+1-1}=\sqrt {3}.

解析

原式=[2a−2(a+1)(a−1)+a+2(a+1)(a−1)]⋅a+1a=\left[\dfrac{2a-2}{\left(a+1\right)\left(a-1\right)}+\dfrac{a+2}{\left(a+1\right)\left(a-1\right)}\right]\cdot \dfrac{a+1}{a}

=3a(a+1)(a−1)⋅a+1a=\dfrac{3a}{\left(a+1\right)\left(a-1\right)}\cdot \dfrac{a+1}{a}

=3a−1=\dfrac{3}{a-1}，

当a=2sin⁡60∘+tan⁡45∘=2×32+1=3+1a=2\sin 60^{\circ}+\tan 45^{\circ}=2\times \dfrac{\sqrt {3}}{2}+1=\sqrt {3}+1时，

原式=33+1−1=3=\dfrac{3}{\sqrt {3}+1-1}=\sqrt {3}.

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