Cho hình chóp S.ABCD có đáy ABCD là

* Hướng dẫn giải

Chọn A.

Cách 1.

Ta có $V_1=V_{S.MNPQ}=V_{S.MNQ}+V_{S.PNQ}$

Ta có $\left\{\begin{array}{l}\left(\alpha \right)\cap \left(SBC\right)=PQ\\ MN//BC\\ MN\subset \left(\alpha \right)\\ BC\subset \left(SBC\right)\end{array}\right.\Rightarrow PQ//MN//BC\Rightarrow \frac{SP}{SC}=\frac{SQ}{SB}=x.$

Có $\frac{V_{S.MNQ}}{V_{S.ADB}}=\frac{SM}{SA}.\frac{SN}{SD}.\frac{SQ}{SB}=\frac{2}{3}.\frac{2}{3}x=\frac{4}{9x}\Rightarrow V_{S.MNQ}=\frac{4x}{9}{V}_{S.ADB}=\frac{4x}{9}.\frac{V}{2}=\frac{2x}{9}V.$

Đồng thời $$\begin{gathered} \frac{V_{S.PNQ}}{V_{S.CDB}}=\frac{SP}{SC}.\frac{SN}{SD}.\frac{SQ}{SB}=x.\frac{2}{3}.x=\frac{2x^2}{3}\Rightarrow V_{S.PNQ}=\frac{2x^2}{3}.V_{S.CDB}=\frac{2x^2}{3}.\frac{V}{2} \\ =\frac{x^2}{3}V. \end{gathered}$$

Như vậy $V_1=\left(\frac{x^2}{3}+\frac{2x}{9}\right)V.$ Mà theo giả thiết ta có $V_1=\frac{1}{2}V$ nên ta suy ra:

$\frac{x^2}{3}+\frac{2x}{9}=\frac{1}{2}\iff \left[\begin{array}{l}x=\frac{-2+\sqrt{58}}{6}\left(Nhan\right)\\ x=\frac{-2-\sqrt{58}}{6}\left(Loai\right)\end{array}\right..$ Vậy $x=\frac{-2+\sqrt{58}}{6}.$

Cách 2:

Đặt $a=\frac{SM}{SA}=\frac{2}{3};b=\frac{SN}{SD}=\frac{2}{3};c=\frac{SP}{SC}.$ Ta có $\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{x}\Rightarrow c=x.$

Lại có $\frac{V_1}{V}=\frac{abcx}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{x}\right)=\frac{x^2}{9}\left(3+\frac{2}{x}\right).$

Mà $\frac{V_1}{V}=\frac{1}{2}\Rightarrow 6x^3+4x^2-9x=0\iff \left[\begin{array}{l}x=0\left(Loai\right)\\ x=\frac{-2+\sqrt{58}}{6}\left(Nhan\right)\\ x=\frac{-2-\sqrt{58}}{6}\left(Loai\right)\end{array}\right..$

Vậy $x=\frac{-2+\sqrt{58}}{6}.$