Đáp án:
a) \(\left[ \begin{array}{l}
x = \sqrt 5 - 1\\
x = - \sqrt 5 - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = - 1\\
Pt \to {x^2} + 2x - 4 = 0\\
\to {x^2} + 2x + 1 = 5\\
\to {\left( {x + 1} \right)^2} = 5\\
\to \left| {x + 1} \right| = \sqrt 5 \\
\to \left[ \begin{array}{l}
x + 1 = \sqrt 5 \\
x + 1 = - \sqrt 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \sqrt 5 - 1\\
x = - \sqrt 5 - 1
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm trái dấu
\(\begin{array}{l}
\to 1.\left( { - {m^2} + m - 2} \right) < 0\\
\to - {m^2} + m - 2 < 0\\
\to {m^2} - m + 2 > 0\\
\to {m^2} - 2m.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4} > 0\\
\to {\left( {m - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} > 0\left( {ld} \right)\forall m\\
\to dpcm
\end{array}\)
c) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to {m^2} - 2m + 1 - 4\left( { - {m^2} + m - 2} \right) \ge 0\\
\to {m^2} - 2m + 1 + 4{m^2} - 4m + 8 \ge 0\\
\to 5{m^2} - 6m + 9 \ge 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 1\\
{x_1}{x_2} = - {m^2} + m - 2
\end{array} \right.\\
A = {x_1}^2 + {x_2}^2\\
= {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {m^2} - 2m + 1 - 2\left( { - {m^2} + m - 2} \right)\\
= {m^2} - 2m + 1 + 2{m^2} - 2m + 4\\
= 3{m^2} - 4m + 5\\
= 3{m^2} - 2.m\sqrt 3 .\dfrac{2}{{\sqrt 3 }} + \dfrac{4}{3} + \dfrac{{11}}{3}\\
= {\left( {m\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{{11}}{3}\\
Do:{\left( {m\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} \ge 0\forall m\\
\to {\left( {m\sqrt 3 - \dfrac{2}{{\sqrt 3 }}} \right)^2} + \dfrac{{11}}{3} \ge \dfrac{{11}}{3}\\
\to Min = \dfrac{{11}}{3}\\
\Leftrightarrow m\sqrt 3 - \dfrac{2}{{\sqrt 3 }} = 0\\
\to m = \dfrac{2}{3}
\end{array}\)
