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九年级数学解答题一般
题目
先化简,再求代数式1a−2÷(1+1a−2\frac{1}{{a-2}}\div (1+\frac{1}{{a-2}}）的值,其中a=2sin⁡60∘+3tan⁡30∘a=2\sin 60^{\circ}+\sqrt{3}\tan 30^{\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=1a−2÷(a−2a−2+1a−2)=\frac{1}{a-2}\div (\frac{a-2}{a-2}+\frac{1}{a-2})
=1a−2÷a−1a−2=\frac{1}{a-2}\div \frac{a-1}{a-2}
=1a−2⋅a−2a−1=\frac{1}{a-2}\cdot \frac{a-2}{a-1}
=1a−1=\frac{1}{a-1}，
当a=2sin⁡60∘+3tan⁡30∘=2×32+3×33=3+1a=2\sin 60^{\circ}+\sqrt{3}\tan 30^{\circ}=2\times \frac{\sqrt{3}}{2}+\sqrt{3}\times \frac{\sqrt{3}}{3}=\sqrt{3}+1时，
原式=13+1−1=13=33=\frac{1}{\sqrt{3}+1-1}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.

解析

原式=1a−2÷(a−2a−2+1a−2)=\frac{1}{a-2}\div (\frac{a-2}{a-2}+\frac{1}{a-2})
=1a−2÷a−1a−2=\frac{1}{a-2}\div \frac{a-1}{a-2}
=1a−2⋅a−2a−1=\frac{1}{a-2}\cdot \frac{a-2}{a-1}
=1a−1=\frac{1}{a-1}，
当a=2sin⁡60∘+3tan⁡30∘=2×32+3×33=3+1a=2\sin 60^{\circ}+\sqrt{3}\tan 30^{\circ}=2\times \frac{\sqrt{3}}{2}+\sqrt{3}\times \frac{\sqrt{3}}{3}=\sqrt{3}+1时，
原式=13+1−1=13=33=\frac{1}{\sqrt{3}+1-1}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.

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