Đáp án:
$a)\left[\begin{array}{l}x=\dfrac{-1+\sqrt{5}}{4} \\x=\dfrac{-1-\sqrt{5}}{4} \end{array} \right.$
$b) min_{a(x)}=-\dfrac{5}{4}$ tại $x=-\dfrac{1}{4}$
Giải thích các bước giải:
$a)a(x)=4x^2+2x-1\\ =4x^2+x+x+\dfrac{1}{4}-\dfrac{5}{4}\\ =2x\left(2x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(2x+\dfrac{1}{2}\right)-\dfrac{5}{4}\\ =\left(2x+\dfrac{1}{2}\right)^2-\dfrac{5}{4} \\ a(x)=0\\ \Leftrightarrow \left(2x+\dfrac{1}{2}\right)^2-\dfrac{5}{4} =0\\ \Leftrightarrow \left(2x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\\ \Leftrightarrow \left[\begin{array}{cc}2x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2} & x \ge-\dfrac{1}{4} \\ 2x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2} & x <-\dfrac{1}{4}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}x=\dfrac{-1+\sqrt{5}}{4} \\x=\dfrac{-1-\sqrt{5}}{4} \end{array} \right.\\ b)a(x)=4x^2+2x-1\\ =\left(2x+\dfrac{1}{2}\right)^2-\dfrac{5}{4} \ge -\dfrac{5}{4} \,\forall \, x$
Dấu "=" xảy ra $\Leftrightarrow x=-\dfrac{1}{4}$
