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

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

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

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

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

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

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

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

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