Đáp án:
$P = 1 + a$
Giải thích các bước giải:
$\quad P = \dfrac{a^{\tfrac13}\left(a^{\tfrac12} - a^{\tfrac52}\right)}{a^{\tfrac14}\left(a^{\tfrac72} - a^{\tfrac{19}{12}}\right)}$
$\to P = \dfrac{a^{\tfrac13}.a^{\tfrac12}\left(1 - a^2\right)}{a^{\tfrac14}.a^{\tfrac{7}{12}}\left(1- a\right)}$
$\to P = \dfrac{a^{\tfrac56}\left(1 - a\right)(1+a)}{a^{\tfrac56}\left(1- a\right)}$
$\to P = 1 + a$
