[Toán 10] Tìm gtnn

E

eye_smile

$M=\dfrac{3}{4}+\dfrac{1}{4}.\dfrac{x^3-x^2-x+1}{x^3+3x^2+3x+1}$

$=\dfrac{3}{4}+\dfrac{1}{4}.\dfrac{(x+1)(x-1)^2}{(x+1)^3}$

$=\dfrac{3}{4}+\dfrac{1}{4}.\dfrac{(x-1)^2}{(x+1)^2} \ge \dfrac{3}{4}$

Dấu = xảy ra \Leftrightarrow $x=1$
 
You must log in or register to reply here.
Top Bottom