Giá trị nhỏ nhất

S

soccan

$A=2x^2+2x+1\\
=2(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4})+\dfrac{1}{2}\\
=2(x+\dfrac{1}{2})^2+2 \ge \dfrac{1}{2}\\
min_A=\dfrac{1}{2} \Longleftrightarrow x=\dfrac{-1}{2}$
 
You must log in or register to reply here.
Top Bottom