$m^2+m=\dfrac{1}{4}$
$↔m^2+2.m.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{4}=0$
$↔\Big(m+\dfrac{1}{2}\Big)^2-\dfrac{1}{2}=0$
$↔\Big(m+\dfrac{1}{2}\Big)^2=\dfrac{1}{2}$
\(↔\left[ \begin{array}{l}m+\dfrac{1}{2}=\dfrac{1}{\sqrt{2}}\to m=\dfrac{\sqrt{2}-1}{2}\\m+\dfrac{1}{2}=-\dfrac{1}{\sqrt{2}}\to m=-\dfrac{1+\sqrt{2}}{2}\end{array} \right.\)
Với $m=\dfrac{\sqrt{2}-1}{2}$
$\to 2.\Big(\dfrac{\sqrt{2}-1}{2}\Big)^4+20.\Big(\dfrac{\sqrt{2}-1}{2}\Big)^3+13.\Big(\dfrac{\sqrt{2}-1}{2}\Big)^2$
$=\dfrac{3}{4}$
Với $m=-\dfrac{1+\sqrt{2}}{2}$
$\to 2.\Big(-\dfrac{1+\sqrt{2}}{2}\Big)+20.\Big(-\dfrac{1+\sqrt{2}}{2}\Big)+13.\Big(-\dfrac{1+\sqrt{2}}{2}\Big)$
$=\dfrac{3}{4}$
Vậy giá trị của biểu thức là $\dfrac{3}{4}$ khi $m^2+m=\dfrac{1}{4}$
