x^2-left(2m+3right)x+m^2+3m+20.left{x^2-left(2m+3right)x+frac{left(2m+3right)^2}{4}right}frac{left(2m+3right)^2+4m^2+12m+8}{4}left(x-frac{2m+3}{2}right)^2frac{8m^2+24m+17}{4}Leftrightarrowhept{begin{cases}2x-2m+3sqrt{8m^2+24m+17}2x-2m+3-sqrt{8m^2+24m+17}end{cases}}để căn có nghĩa thì8m^2+24m+17left(m^2+3m+frac{9}{4}right)-frac{1}{8}ge0left(m+frac{3}{2}right)^2gefrac{1}{8} suy ra m.....vậy pt có 2 nghiệm phân biệt với m.....Leftrightarrowhept{begin{cases}x1frac{1}{2}sqrt{8m^2+24+17}+m-frac{3}{2}...
Đọc tiếp
\(x^2-\left(2m+3\right)x+m^2+3m+2=0.\)
\(\left\{x^2-\left(2m+3\right)x+\frac{\left(2m+3\right)^2}{4}\right\}=\frac{\left(2m+3\right)^2+4m^2+12m+8}{4}\)
\(\left(x-\frac{2m+3}{2}\right)^2=\frac{8m^2+24m+17}{4}\)
\(\Leftrightarrow\hept{\begin{cases}2x-2m+3=\sqrt{8m^2+24m+17}\\2x-2m+3=-\sqrt{8m^2+24m+17}\end{cases}}\)
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 =))
