Rút gọn biểu thức P=({log_a}b+{log_b}a+2)({log_a}b-{log_ab}b){log_b}a-1

$\begin{array}{l} ({\log _a}b + {\log _b}a + 2)\left( {{{\log }_a}b - {{\log }_{ab}}b} \right){\log _b}a - 1\\ = ({\log _a}b + {\log _b}a + 2)\left( {{{\log }_a}b - \frac{{{{\log }_a}b}}{{{{\log }_a}(ab)}}} \right){\log _b}a - 1\\ = ({\log _a}b + \frac{1}{{{{\log }_a}b}} + 2)\left( {1 - \frac{1}{{1 + {{\log }_a}b}}} \right) - 1 \end{array}$

Đặt $t = {\log _a}b$

$\Rightarrow \left( {1 + \frac{1}{t} + 2} \right)\left( {1 - \frac{1}{{1 + t}}} \right) - 1 = \frac{{{t^2} + 2t + 1}}{t}.\frac{t}{{t + 1}} - 1 = t = {\log _a}b$