Cho \({\log _8}\left| x \right| + {\log _4}{y^2} =...

B

- Hướng dẫn giải

Điều kiện: $x,y \ne 0$

Theo đề bài ta có hệ phương trình:

$\begin{array}{l} \left\{ \begin{array}{l} {\log _8}\left| x \right| + {\log _4}{y^2} = 5\\ {\log _8}\left| y \right| + {\log _4}{x^2} = 7 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \frac{1}{3}{\log _2}\left| x \right| + {\log _2}\left| y \right| = 5\\ \frac{1}{3}{\log _2}\left| y \right| + {\log _2}\left| x \right| = 7 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} {\log _2}\left| x \right| + {\log _2}{\left| y \right|^3} = 15\\ {\log _2}\left| y \right| + {\log _2}{\left| x \right|^3} = 21 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\log _2}\left| {x{y^3}} \right| = 15\\ {\log _2}\left| {{x^3}y} \right| = 21 \end{array} \right.\\  \Leftrightarrow \left\{ \begin{array}{l} \left| {x{y^3}} \right| = {2^{15}}{\rm{ (*)}}\\ \left| {{x^3}y} \right| = {2^{21}} \end{array} \right. \Leftrightarrow \frac{{\left| {{x^3}y} \right|}}{{\left| {x{y^3}} \right|}} = 64 \Leftrightarrow {\left| {\frac{x}{y}} \right|^2} = 64 \Leftrightarrow \left| {\frac{x}{y}} \right| = 8 \Leftrightarrow \left| x \right| = 8\left| y \right| \end{array}$

Thay vào (*) ta có $8{y^4} = {2^{15}} \Leftrightarrow \left| y \right| = \sqrt[4]{{4096}} = 8$

Khi đó ta có $P = \left| x \right| - \left| y \right| = 8\left| y \right| - \left| y \right| = 7\left| y \right| = 7.8 = 56$