Cho a , b , x > 0 ; a > b v \` a b , x ≠ 1 thỏa mãn log x (a + 2b)/ 3 = log x √ a + 1 log b x^2 .

Đáp án: \(\frac{5}{4}\).

\[{\log _x}\frac{{a + 2b}}{3} = {\log _x}\sqrt a  + \frac{1}{{{{\log }_b}{x^2}}} \Leftrightarrow {\log _x}\frac{{a + 2b}}{3} = {\log _x}\sqrt a  + {\log _x}\sqrt b \]

\[ \Leftrightarrow a + 2b = 3\sqrt {ab}  \Leftrightarrow {a^2} - 5ab + 4{b^2} = 0 \Leftrightarrow \left( {a - b} \right)\left( {a - 4b} \right) = 0 \Leftrightarrow a = 4b\] (do \(a > b\)).

\[P = \frac{{2{a^2} + 3ab + {b^2}}}{{{{(a + 2b)}^2}}} = \frac{{32{b^2} + 12{b^2} + {b^2}}}{{36{b^2}}} = \frac{5}{4}\].