Giải thích các bước giải:
Ta có :
$B=\dfrac{x^2-x+7}{x^2-x-1}$
$\to \dfrac{1}{B}=\dfrac{x^2-x-1}{x^2-x+7}$
$\to \dfrac{1}{B}+\dfrac{5}{27}=\dfrac{x^2-x-1}{x^2-x+7}+\dfrac{5}{27}$
$\to \dfrac{1}{B}+\dfrac{5}{27}=\dfrac{27(x^2-x-1)+5(x^2-x+7)}{27(x^2-x+7)}$
$\to \dfrac{1}{B}+\dfrac{5}{27}=\dfrac{8(2x-1)^2}{27(x^2-x+7)}\ge 0\quad\forall x$
$\to \dfrac{1}{B}\ge -\dfrac{5}{27}$
$\to B\le -\dfrac{27}{5}\to x=\dfrac12$