Question

7．計(jì)算：$\frac{1+\sqrt{2015}（\sqrt{2014}-\sqrt{2013}）}{\sqrt{2013}+\sqrt{2014}+\sqrt{2015}}$+$\sqrt{2013}$．

Answer 1

分析 設(shè)a=2014，則原式=$\frac{1+\sqrt{a+1}（\sqrt{a}-\sqrt{a-1}）}{\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}$+$\sqrt{a-1}$，再進(jìn)行通分，把分子進(jìn)行合并，然后把分子分解因式后約分即可．

解答 解：設(shè)a=2014，
原式=$\frac{1+\sqrt{a+1}（\sqrt{a}-\sqrt{a-1}）}{\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}$+$\sqrt{a-1}$
=$\frac{1+\sqrt{a}•\sqrt{a+1}-\sqrt{a-1}•\sqrt{a+1}+a-1+\sqrt{a}•\sqrt{a-1}+\sqrt{a-1}•\sqrt{a+1}}{\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}$
=$\frac{a+\sqrt{a}•\sqrt{a-1}+\sqrt{a}•\sqrt{a+1}}{\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}$
=$\frac{\sqrt{a}（\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}）}{\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}$
=$\sqrt{a}$
=$\sqrt{2014}$．

點(diǎn)評(píng) 本題考查了二次根式的計(jì)算：先把各二次根式化為最簡(jiǎn)二次根式，再進(jìn)行二次根式的乘除運(yùn)算，然后合并同類二次根式．在二次根式的混合運(yùn)算中，如能結(jié)合題目特點(diǎn)，靈活運(yùn)用二次根式的性質(zhì)，選擇恰當(dāng)?shù)慕忸}途徑，往往能事半功倍．