Đáp án:
b) \(\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
(m - 1){x^2} - 2(m - 1){x^2} + 2mx + m + 1 = 0\\
\to \left( {m - 1 - 2m + 2} \right){x^2} + 2mx + m + 1 = 0\\
\to \left( {1 - m} \right){x^2} + 2mx + m + 1 = 0\\
a)Thay:m = 4\\
Pt \to - 3{x^2} + 8x + 5 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{4 + \sqrt {31} }}{3}\\
x = \dfrac{{4 - \sqrt {31} }}{3}
\end{array} \right.\\
b)DK:\left( {1 - m} \right)\left( {m + 1} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - m > 0\\
m + 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - m < 0\\
m + 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 > m\\
m < - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
m > 1\\
m > - 1
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
c)A = {x_1}^2{x_2} + {x_2}^2{x_1}\\
= {x_1}{x_2}\left( {{x_1} + {x_2}} \right)\\
= \dfrac{{m + 1}}{{1 - m}}.\dfrac{{ - 2m}}{{1 - m}}\\
= \dfrac{{ - 2{m^2} - 2m}}{{{{\left( {1 - m} \right)}^2}}}
\end{array}\)
