\(log_{ab}b=\dfrac{1}{log_bab}=\dfrac{1}{log_ba+1}=\dfrac{1}{\dfrac{1}{log_ab}+1}=\dfrac{log_ab}{log_ab+1}\)
Đặt \(log_ab=x\) (cho gọn biểu thức)
\(P=\left(x+\dfrac{1}{x}+2\right)\left(x-\dfrac{x}{x+1}\right).\dfrac{1}{x}-1\)
\(=\left(\dfrac{x^2+2x+1}{x}\right)\left(\dfrac{x}{x+1}\right).\dfrac{1}{x}-1\)
\(=\dfrac{x+1}{x}-1=\dfrac{1}{x}=log_ba\)