Question

9．$\frac{7}{3}$和它的倒數(shù)的積加上$\frac{3}{5}$等于1$\frac{3}{5}$，圓有無(wú)數(shù)條對(duì)稱(chēng)軸．

Answer 1

分析 $\frac{7}{3}$的倒數(shù)是$\frac{3}{7}$，再算$\frac{7}{3}$與$\frac{3}{7}$的積，所得的積加上$\frac{3}{5}$即可求解；根據(jù)圓的性質(zhì)和對(duì)稱(chēng)軸的定義即可求解．

解答 解：$\frac{7}{3}$的倒數(shù)是$\frac{3}{7}$
$\frac{7}{3}$×$\frac{3}{7}$+$\frac{3}{5}$
=1+$\frac{3}{5}$
=1$\frac{3}{5}$
答：$\frac{7}{3}$和它的倒數(shù)的積加上$\frac{3}{5}$等于1$\frac{3}{5}$，圓有無(wú)數(shù)條對(duì)稱(chēng)軸．
故答案為：1$\frac{3}{5}$，無(wú)數(shù)．

點(diǎn)評(píng) 根據(jù)題意，先弄清運(yùn)算順序，然后再列式進(jìn)行解答．同時(shí)考查了倒數(shù)的定義，圓的性質(zhì)和對(duì)稱(chēng)軸的定義．