a. Ta có: \(\Delta>0\Rightarrow8^2-4\left(4-8m\right)>0\Leftrightarrow m< -1,5\)
Theo định lí Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=8\\x_1x_2=4-8m\end{matrix}\right.\)
Vì \(1< x_1< x_2\) nên \(x_1x_2>1\Leftrightarrow4-8m>1\Leftrightarrow3>8m\Leftrightarrow m< \dfrac{3}{8}\)
Lại có: \(x_1=\dfrac{8-\sqrt{48+32m}}{2}>1\)
\(\Leftrightarrow8-\sqrt{48+32m}>2\)
\(\Leftrightarrow\sqrt{48+32m}< 6\)
\(\Leftrightarrow48+32m< 36\)
\(\Leftrightarrow m< -\dfrac{3}{8}\)
Vậy \(-\dfrac{3}{2}< m< \dfrac{-3}{8}\) thì phương trình thỏa mãn điều kiện
\(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(mà:\left(a-b\right)^2;\left(b-c\right)^2;\left(c-a\right)^2\ge0\Rightarrow dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)
\(a;\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left(x1-1\right)\left(x2-1\right)>0\\x1+x2-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}8^2-4\left(4-8m\right)>0\\x1x2-\left(x1+x2\right)+1>0\\8-2=6>0\left(đúng\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{3}{2}\\-8m-8+1>0\Leftrightarrow m< -\dfrac{7}{8}\end{matrix}\right.\)\(\Rightarrow-\dfrac{3}{2}< m< -\dfrac{7}{8}\)
\(b;a^2+b^2+c^2=ab+bc+ca\Leftrightarrow a=b=c\)
\(\Rightarrow a+b-c=\sqrt{3}\Leftrightarrow a=b=c=\sqrt{3}\)
\(\Rightarrow A=\sqrt{3+1}+3\sqrt{3}^2=11\)
