Cho \({\rm{log}}_2^2\left( {xy} \right) = {\rm{lo}}{{\rm{g}}_2}\left( {\frac{x}{4}} \right){\rm{lo}}{{\rm{g}}_2}\left( {4y} \right)\). Biểu thức \(P = {\rm{lo}}{{\rm{g}}_3}\left( {x + 4y + 4} \right) + {\rm{lo}}{{\rm{g}}_2}\left( {x - 4y - 1} \right)\) có giá trị bằng bao nhiêu?

Giải thích

Điều kiện: \(\left\{ {\begin{array}{*{20}{l}}{x > 0}\\{y > 0}\\{x - 4y - 1 > 0}\end{array}} \right.\)

Ta có \({\rm{log}}_2^2\left( {xy} \right) = {\rm{lo}}{{\rm{g}}_2}\left( {\frac{x}{4}} \right){\rm{lo}}{{\rm{g}}_2}\left( {4y} \right) \Leftrightarrow {\left( {{\rm{lo}}{{\rm{g}}_2}x + {\rm{lo}}{{\rm{g}}_2}y} \right)^2} = \left( {{\rm{lo}}{{\rm{g}}_2}x - 2} \right)\left( {{\rm{lo}}{{\rm{g}}_2}y + 2} \right){\rm{\;}}\) (1).

Đặt \({\rm{lo}}{{\rm{g}}_2}x = a;{\rm{lo}}{{\rm{g}}_2}y = b\), ta có (1) trở thành :

\({(a + b)^2} = \left( {a - 2} \right)\left( {b + 2} \right) \Leftrightarrow {a^2} + ab - 2a + {b^2} + 2b + 4 = 0\)

\( \Leftrightarrow 2{a^2} + 2ab - 4a + 2{b^2} + 4b + 8 = 0 \Leftrightarrow {(a + b)^2} + {(a - 2)^2} + {(b + 2)^2} = 0\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a + b = 0}\\{a - 2 = 0}\\{b + 2 = 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a = 2}\\{b =  - 2}\end{array}} \right.} \right.\)

Với \(\left\{ {\begin{array}{*{20}{l}}{a = 2}\\{b =  - 2}\end{array}} \right.\), ta có \(\left\{ {\begin{array}{*{20}{l}}{{\rm{lo}}{{\rm{g}}_2}x = 2}\\{{\rm{lo}}{{\rm{g}}_2}y =  - 2}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x = 4}\\{y = \frac{1}{4}}\end{array}} \right.} \right.\) (thỏa mãn điều kiện).

Khi đó \(P = {\rm{lo}}{{\rm{g}}_3}\left( {4 + 4.\frac{1}{4} + 4} \right) + {\rm{lo}}{{\rm{g}}_2}\left( {4 - 4.\frac{1}{4} - 1} \right) = 3\).

 Chọn A