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Câu hỏi: Cho hình lập phương $ABCD.A'B'C'D'.$ Gọi $M$ là trung điểm của $AA'$ và $N$ là điểm nằm trên cạnh $DD'$ sao cho $DN=3ND'.$ Mặt phẳng $\left( BMN \right)$ chia khối lập phương thành hai phần có thể tích lần lượt là ${{V}_{1}},{{V}_{2}}\left( {{V}_{1}}<{{V}_{2}} \right)$, tính $\dfrac{{{V}_{1}}}{{{V}_{2}}}.$
A. $\dfrac{3}{5}$.
B. $\dfrac{5}{11}$.
C. $\dfrac{3}{8}$.
D. $\dfrac{3}{13}$.
Gọi $S=MN\cap AD$, $E=SB\cap DC$ và $E=NF\cap DC$.
Ta có $\dfrac{MA}{ND}=\dfrac{\dfrac{1}{2}AA'}{\dfrac{3}{4}DD'}=\dfrac{2}{3}\Rightarrow \dfrac{SA}{SD}=\dfrac{SB}{SE}=\dfrac{SM}{SN}=\dfrac{MA}{ND}=\dfrac{2}{3}$ và $\dfrac{EF}{EN}=\dfrac{EC}{ED}=\dfrac{EB}{ES}=\dfrac{ES-SB}{ES}=\dfrac{1}{3}$.
Ta có ${{V}_{MAB.NDCF}}={{V}_{SNDE}}-{{V}_{SMAB}}-{{V}_{BCFE}}={{V}_{SNDE}}-\dfrac{SM}{SN}\dfrac{SA}{SD}\dfrac{SB}{SE}{{V}_{SNDE}}-\dfrac{EF}{EN}\dfrac{EB}{ES}\dfrac{EC}{ED}{{V}_{SNDE}}$
$={{V}_{SNDE}}-{{\left( \dfrac{SM}{SN} \right)}^{3}}{{V}_{SNDE}}-{{\left( \dfrac{EF}{EN} \right)}^{3}}{{V}_{SNDE}}={{V}_{SNDE}}-{{\left( \dfrac{2}{3} \right)}^{3}}{{V}_{SNDE}}-{{\left( \dfrac{1}{3} \right)}^{3}}{{V}_{SNDE}}=\dfrac{2}{3}{{V}_{SNDE}}$.
Mà ${{V}_{SNDE}}=\dfrac{1}{3}ND.{{S}_{SDE}}=\dfrac{1}{6}ND.DS.DE=\dfrac{1}{6}\dfrac{3}{4}DD'.3DA.\dfrac{3}{2}DC=\dfrac{9}{16}{{V}_{ABCD.A'B'C'D'}}$
$\Rightarrow {{V}_{MAB.NDCF}}=\dfrac{2}{3}{{V}_{SNDE}}=\dfrac{3}{8}{{V}_{ABCD.A'B'C'D'}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{3}{5}$.
A. $\dfrac{3}{5}$.
B. $\dfrac{5}{11}$.
C. $\dfrac{3}{8}$.
D. $\dfrac{3}{13}$.
Ta có $\dfrac{MA}{ND}=\dfrac{\dfrac{1}{2}AA'}{\dfrac{3}{4}DD'}=\dfrac{2}{3}\Rightarrow \dfrac{SA}{SD}=\dfrac{SB}{SE}=\dfrac{SM}{SN}=\dfrac{MA}{ND}=\dfrac{2}{3}$ và $\dfrac{EF}{EN}=\dfrac{EC}{ED}=\dfrac{EB}{ES}=\dfrac{ES-SB}{ES}=\dfrac{1}{3}$.
Ta có ${{V}_{MAB.NDCF}}={{V}_{SNDE}}-{{V}_{SMAB}}-{{V}_{BCFE}}={{V}_{SNDE}}-\dfrac{SM}{SN}\dfrac{SA}{SD}\dfrac{SB}{SE}{{V}_{SNDE}}-\dfrac{EF}{EN}\dfrac{EB}{ES}\dfrac{EC}{ED}{{V}_{SNDE}}$
$={{V}_{SNDE}}-{{\left( \dfrac{SM}{SN} \right)}^{3}}{{V}_{SNDE}}-{{\left( \dfrac{EF}{EN} \right)}^{3}}{{V}_{SNDE}}={{V}_{SNDE}}-{{\left( \dfrac{2}{3} \right)}^{3}}{{V}_{SNDE}}-{{\left( \dfrac{1}{3} \right)}^{3}}{{V}_{SNDE}}=\dfrac{2}{3}{{V}_{SNDE}}$.
Mà ${{V}_{SNDE}}=\dfrac{1}{3}ND.{{S}_{SDE}}=\dfrac{1}{6}ND.DS.DE=\dfrac{1}{6}\dfrac{3}{4}DD'.3DA.\dfrac{3}{2}DC=\dfrac{9}{16}{{V}_{ABCD.A'B'C'D'}}$
$\Rightarrow {{V}_{MAB.NDCF}}=\dfrac{2}{3}{{V}_{SNDE}}=\dfrac{3}{8}{{V}_{ABCD.A'B'C'D'}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{3}{5}$.
Đáp án A.