Đáp án:
$\dfrac{3}{25x^2-7x+4}\le \dfrac{100}{117}$
Giải thích các bước giải:
$A=\dfrac{3}{25x^2-7x+4}$
$\text{Ta có: }25x^2-7x+4=(5x)^2-2.5x.\dfrac{7}{10}+(\dfrac{7}{10})^2+(4-(\dfrac{7}{10})^2)$
$=(5x-\dfrac{7}{10})^2+\dfrac{351}{100}\ge 0+\dfrac{351}{100}=\dfrac{351}{100}\quad\forall x$
$\rightarrow A=\dfrac{3}{25x^2-7x+4}\le \dfrac{3}{\dfrac{351}{100}}=\dfrac{100}{117}$
$\text{Dấu = xảy ra }\leftrightarrow 5x-\dfrac{7}{10}=0\leftrightarrow x=\dfrac{7}{50}$
