\(x^2-\left(2m+3\right)x+m^2+3m+2=0.\)

\(\left\{x^2-\left(2m+3\right)x+\frac{\left(2m+3\right)^2}{4}\right\}=\frac{\left(2m+3\right)^2+4m^2+12m+8}{4}\)

\(\left(x-\frac{2m+3}{2}\right)^2=\frac{8m^2+24m+17}{4}\)

\(\Leftrightarrow\hept{\begin{cases}2x-2m+3=\sqrt{8m^2+24m+17}\\2x-2m+3=-\sqrt{8m^2+24m+17}\end{cases}}\)

để căn có nghĩa thì

\(8m^2+24m+17=\left(m^2+3m+\frac{9}{4}\right)-\frac{1}{8}\ge0\)

\(\left(m+\frac{3}{2}\right)^2\ge\frac{1}{8}\) " suy ra m.....

vậy pt có 2 nghiệm phân biệt với m.....

\(\Leftrightarrow\hept{\begin{cases}x1=\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\\x2=-\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\end{cases}}\)

\(x1< -3\Leftrightarrow-3< \frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\)

\(\Leftrightarrow m>-3-\frac{1}{2}\sqrt{8m^2+24+17}+\frac{3}{2}\)

\(x1< x2\Leftrightarrow\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}< -\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}\)

\(\Leftrightarrow0< -\sqrt{8m^2+24+17}\)

\(x2< 6\Leftrightarrow-\frac{1}{2}\sqrt{8m^2+24+17}+m-\frac{3}{2}< 6\)

\(\Leftrightarrow m< 6+\frac{1}{2}\sqrt{8m^2+24+17}+\frac{3}{2}\)

dcpcm =))

ta có : 4x-2y=-6

\(\Leftrightarrow-4\frac{2m^2+8m+38}{9m+15}-2\frac{3m^2-24m-3}{9m+15}=-6\)

\(\Leftrightarrow-4\left(2m^2+8m+38\right)-2\left(3m^2-24-3\right)=-6\left(9m+15\right)\)

\(\Leftrightarrow-8m^2-32m-152-6m^2+48m+6=-54m-90\)

\(\Leftrightarrow-8m^2-32m-152-6m^2+48m+6+54m+90=0\)

\(\Leftrightarrow-14m^2+70m-56=0\)

\(\Leftrightarrow-7\left(2m^2-10m+8\right)=0\)

\(\Leftrightarrow2m^2-10m+8=0\)

\(\Leftrightarrow2m^2-2m-8m+8=0\)

\(\Leftrightarrow\left(2m-8\right)\left(m-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2m-8=0\\m-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=4\\m=1\end{cases}}\)