Áp dụng sơ đồ đường chéo ta có:
Do đó: \(\dfrac{V_{ddA}}{V_{ddB}}=\dfrac{3-C_{M_B}}{C_{M_A}-3}\)
mà \(\dfrac{V_{ddA}}{V_{ddB}}=\dfrac{3}{5}\left(gt\right)\)
nên \(\dfrac{3-C_{M_B}}{C_{M_A}-3}=\dfrac{3}{5}\left(1\right)\)
Lại có: \(C_{M_A}=2.C_{M_B}\left(gt\right)\left(2\right)\)
Suy ra: \(\left(1\right)\Leftrightarrow\dfrac{3-C_{M_B}}{2.C_{M_B}-3}=\dfrac{3}{5}\)
\(\Leftrightarrow5\left(3-C_{M_B}\right)=3\left(2.C_{M_B}-3\right)\)
\(\Leftrightarrow15-5.C_{M_B}=6.C_{M_B}-9\)
\(\Leftrightarrow-5.C_{M_B}-6.C_{M_B}=-9-15\)
\(\Leftrightarrow-11.C_{M_B}=-24\)
\(\Leftrightarrow C_{M_B}=\dfrac{-24}{-11}\approx2,18\left(M\right)\)
Từ (2) \(\Rightarrow C_{M_A}\approx2.2,18\approx4,36\left(M\right)\)
