Đáp án:
-2,4
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{{d_N}'}}{{{d_N}}} = 2 \Rightarrow {d_N}' = 2{d_N}\\
\dfrac{1}{f} = \dfrac{1}{{{d_N}}} + \dfrac{1}{{{d_N}'}} = \dfrac{3}{{2{d_N}}} \Rightarrow {d_N} = \dfrac{3}{2}f\\
\dfrac{{{d_M}'}}{{{d_M}}} = 3 \Rightarrow {d_M}' = 3{d_M}\\
\dfrac{1}{f} = \dfrac{1}{{{d_M}}} + \dfrac{1}{{{d_M}'}} = \dfrac{4}{{3{d_N}}} \Rightarrow {d_N} = \dfrac{4}{3}f\\
{d_P} = \dfrac{{{d_M} + {d_N}}}{2} = \dfrac{{17}}{{12}}f\\
\dfrac{1}{f} = \dfrac{1}{{{d_P}}} + \dfrac{1}{{{d_P}'}} \Rightarrow {d_P}' = \dfrac{{17}}{5}f\\
{k_P} = - \dfrac{{{d_P}'}}{{{d_P}}} = - 2,4
\end{array}\)
