Giải thích các bước giải:
Bài 1:
$y=\dfrac{3x^3+7}{x^2}=3x+\dfrac{7}{x^2}=\dfrac{3x}{2}+\dfrac{3x}{2}+\dfrac{7}{x^2}\ge 3\sqrt[3]{\dfrac{3x}{2}.\dfrac{3x}{2}.\dfrac{7}{x^2}}=3\sqrt[3]{\dfrac{63}{4}}$
Dấu = xảy ra khi $\dfrac{3x}{2}=\dfrac{7}{x^2}\to x=\sqrt[3]{\dfrac{14}{3}}$
Bài 2:
$y=\dfrac{x}{3-2x}+\dfrac{3}{x}$
$\to y=\dfrac{x^2-6x+9}{(3-2x)x}$
$\to y=\dfrac{(x-3)^2}{(3-2x)x}$
$\to |y|=\dfrac{(x-3)^2}{|(3-2x)x|}$
Mà $(3-2x)x\le \dfrac{(3-2x+x)^2}{4}=\dfrac{(x-3)^2}{4}$
$\to \dfrac{(x-3)^2}{|(3-2x)x|}\ge 4$
$\to |y|\ge 4\to y\ge 4\to x=1$
Hoặc $y\le -4\to x=\dfrac{(x-3)^2}{(3-2x)x}=-4\to x=\dfrac{3+6\sqrt{2}}{7}$
