$\begin{array}{l}4)\\\text{- Ta có : $\rm m_1 = 2m_2\\V_1=\dfrac13V_2$}\\\to \rm \dfrac{D_1}{D_2}=\dfrac{\dfrac{m_1}{V_1}}{\dfrac{m_2}{V_2}}=\dfrac{\dfrac{2m_2}{\dfrac13V_2}}{\dfrac{m_2}{V_2}}=\dfrac{6.\dfrac{m_2}{V_2}}{\dfrac{m_2}{V_2}}=6\\\to\rm D_1=6D_2\\\text{- Vậy $\rm D_1=6D_2$}\\\,\\5)\\\text{- Ta có : $\rm m_1=m_2\\D_1=2D_2$}\\\to \rm \dfrac{V_1}{V_2}=\dfrac{\dfrac{m_1}{D_1}}{\dfrac{m_2}{D_2}}=\dfrac{\dfrac{m_2}{2D_2}}{\dfrac{m_2}{D_2}}=\dfrac{\dfrac12.\dfrac{m_2}{D_2}}{\dfrac{m_2}{D_2}}=\dfrac12\\\to\rm V_1=\dfrac12V_2\\\to\rm V_2=2V_1\\\text{- Vậy $\rm V_2=2V_1$} \end{array}$
