Cho \(a,b > 0, a,b \ne 1;\,a \ne {b^2}\).

* Hướng dẫn giải

\(P = {\log _{\sqrt a }}{b^2} + \frac{2}{{{{\log }_{\frac{a}{{{b^2}}}}}a}} = 4{\log _a}b + 2{\log _a}\frac{a}{{{b^2}}} = 4{\log _a}b + 2\left( {{{\log }_a}a - 2{{\log }_a}b} \right) = 2\)