x=1 là nghiệm của pt
$\to$ Phương trình trở thành $\dfrac{1}{9}-m=\dfrac{1}{3}(2m+3)(m-1)$
$↔2m^2+m-3=\dfrac{1}{3}-3m$
$↔2m^2+4m-\dfrac{10}{3}=0$
$↔2m^2+4m+2-\dfrac{16}{3}=0$
$↔2(m+1)^2=\dfrac{16}{3}$
$↔(m+1)^2=\dfrac{16}{6}$
\(\leftrightarrow\left[ \begin{array}{l}m+1=\dfrac{4}{\sqrt{6}}\\m+1=-\dfrac{4}{\sqrt{6}}\end{array} \right.\)
\(\leftrightarrow\left[ \begin{array}{l}m=\dfrac{2\sqrt{6}-3}{3}\\m=-\dfrac{2\sqrt{6}+3}{3}\end{array} \right.\)
