Một hình hộp chữ nhật có kích thước a ...

B

- Hướng dẫn giải

Đáp án B.

Ta có $\left\{\begin{array}{l}V=abc\\ S=2\left(ab+bc+ca\right)\end{array}\right.$. Theo đề ta có:

$\begin{array}{l}abc=2\left(ab+bc+ca\right);\\ 1\le a\le b\le c\\ \iff 1=2.\frac{ab+bc+ca}{abc}\\ \iff \frac{2}{a}+\frac{2}{b}+\frac{2}{c}=1\end{array}$

Ta có $1=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}$$\le \frac{2}{a}+\frac{2}{a}+\frac{2}{a}=\frac{6}{a}$$\Rightarrow a\le 6$. Kết hợp với $\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=1$ ta có:

$a=3\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{6}\Rightarrow 6<b\le 12$

$\Rightarrow \left(a;b;c\right)e\{\left(3;7;42\right),\left(3;8;24\right),$$\left(3;9;18\right),\left(3;10;15\right),\left(3;12;12\right)\}$

Với $a=4\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{4}$$\Rightarrow 4<b\le 8$$\Rightarrow \left(a;b;c\right)\in \left\{\left(4;5;20\right),\left(4;6;12\right),\left(4;8;8\right)\right\}$

với $a=5$$\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{3}{10}<\frac{1}{3}$$\Rightarrow b<6\Rightarrow \left(a;b;c\right)\in \left\{\left(5;5;10\right)\right\}$

với $a=6\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{3}$$\Rightarrow b\le 6\Rightarrow \left(a;b;c\right)\in \left\{\left(6;6;6\right)\right\}$

$\Rightarrow \left(a;b;c\right)\in$$\left\{\begin{array}{l}\left(3;7;42\right),\left(3;8;24\right),\left(3;9;18\right),\\ \left(3;10;15\right),\left(3;12;12\right)\\ \left(4;6;20\right),\left(4;6;12\right),\left(4;8;8\right),\\ \left(5;5;10\right),\left(6;6;6\right)\end{array}\right\}$

Vậy ta chọn B.