Đáp án A
ln 1 - 2 x x + y = 3 x + y - 1 , 0 < x < 1 2 , y > 0 ⇔ ln 1 - 2 x + 1 - 2 x = ln ( x + y ) + x + y f t = t + ln t ⇒ f ' t = 1 + 1 t > 0 ⇒ f 1 - 2 x = f ( x + y ) ⇔ 1 - 2 x = x + y ⇔ y = 1 - 3 x P = 1 x + 1 x y = 1 x + 1 x 1 - 3 x ⇒ P ' = - 1 x 2 + 6 x - 1 2 x 1 - 3 x x ( 1 - 3 x ) = - 2 x 1 - 3 x ( 1 - 3 x ) + ( 6 x - 1 ) x 2 x 1 - 3 x x 2 ( 1 - 3 x ) P ' = 0 ⇔ 2 x 1 - 3 x ( 1 - 3 x ) = 6 x 2 - x ⇔ 6 x 2 - x > 0 4 x ( 1 - 3 x ) 3 = 6 x 2 - x 2 ⇔ [ x < 0 x > 1 6 4 x - 36 x 2 + 108 x 3 - 108 x 4 = 26 x 4 - 12 x 3 + x 2 ⇔ x = y = 1 4 ⇒ P = 8