Đáp án:
a.3
b.36
Giải thích các bước giải:
a.$y=\dfrac{x^2+x+1}{x}=x+1+\dfrac{1}{x}=(x+\dfrac{1}{x})+1\ge 2\sqrt{x.\dfrac{1}{x}}+1=3$
$\rightarrow Miny=3$
$\begin{split}b.y&=\dfrac{18}{x}+\dfrac{36}{4-2x}\\&=(\dfrac{18}{x}+18x)+(\dfrac{36}{4-2x}+9(4-2x))-36\\&\ge 2\sqrt{\dfrac{18}{x}.18x}+2\sqrt{\dfrac{36}{4-2x}9(4-2x)}-36\\&=36\end{split}$
$\rightarrow Min y=36$