《2025年名校课堂九年级数学全一册湘教版广西专版》

答案： 任务1:$\frac{b_{1}}{l_{1}}=\frac{b_{2}}{l_{2}}$.理由如下:$\because P_{1}D_{1}// P_{2}D_{2}$,$\therefore \triangle P_{1}D_{1}O\backsim \triangle P_{2}D_{2}O$.$\therefore \frac{P_{1}D_{1}}{P_{2}D_{2}}=\frac{D_{1}O}{D_{2}O}$,即$\frac{b_{1}}{b_{2}}=\frac{l_{1}}{l_{2}}$.$\therefore \frac{b_{1}}{l_{1}}=\frac{b_{2}}{l_{2}}$.任务2:$\because \frac{b_{1}}{l_{1}}=\frac{b_{2}}{l_{2}}$,且$b_{1}=1.6\ cm$,$b_{2}=1\ cm$,$l_{1}=40\ cm$,$\therefore \frac{1.6}{40}=\frac{1}{l_{2}}$.$\therefore l_{2}=25\ cm$.$\therefore$②号“E”的测量距离$l_{2}$为$25\ cm$.任务3:过点$C$作$CF\perp A'B'$于点$F$,交$MN$于点$E$.由题意,得$A'B'=AB=0.8\ m$,$A'B'// MN$,$\therefore CE\perp MN$,$\triangle CMN\backsim \triangle CA'B'$.$\therefore \frac{CE}{CF}=\frac{MN}{A'B'}$.由题意,得$CF=5\ m$,$EF=3\ m$,$\therefore CE=CF-EF=2\ m$.$\therefore \frac{2}{5}=\frac{MN}{0.8}$.$\therefore MN=0.32\ m$.$\therefore$镜长至少为$0.32\ m$.