Question

16．化簡(jiǎn)$\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+2\sqrt{2}+1}$．

Answer 1

分析 可以先求出所求式子的倒數(shù)的值，然后再求出所求式子的值，本題得以解決．

解答 解：∵$\frac{\sqrt{3}+2\sqrt{2}+1}{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}$
=$\frac{\sqrt{3}+\sqrt{2}+\sqrt{2}+1}{\sqrt{3}（\sqrt{2}+1）+\sqrt{2}（\sqrt{2}+1）}$
=$\frac{（\sqrt{3}+\sqrt{2}）+（\sqrt{2}+1）}{（\sqrt{2}+1）（\sqrt{3}+\sqrt{2}）}$
=$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}$
=$\sqrt{2}-1+\sqrt{3}-\sqrt{2}$
=$\sqrt{3}-1$，
∴$\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+2\sqrt{2}+1}$=$\frac{1}{\sqrt{3}-1}$=$\frac{\sqrt{3}+1}{2}$．

點(diǎn)評(píng) 本題考查實(shí)數(shù)的運(yùn)算，解題的關(guān)鍵是明確題意，可以發(fā)現(xiàn)先求所求式子的倒數(shù)的值，再通過(guò)轉(zhuǎn)化的思想解答．