Đáp án:
a. \(\left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = 2\\
Pt \to {x^2} + 2x = 0\\
\to x\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
b.Xét:\\
Δ'= {m^2} - 4\left( {m - 2} \right)\\
= {m^2} - 4m + 8\\
= {m^2} - 4m + 4 + 4\\
= {\left( {m - 2} \right)^2} + 4 > 0\left( {ld} \right)\forall m \in R
\end{array}\)
⇒ Phương trình luôn có 2 nghiệm phân biệt với mọi m
c. Xét:
\(\begin{array}{l}
\frac{{{{\left( {{x_1}{x_2}} \right)}^2} - 2\left( {{x_1}^2 + {x_2}^2} \right) + 4}}{{{x_1}{x_2} + \left( {{x_1} + {x_2}} \right) + 1}} = 4\\
\to \frac{{{{\left( {{x_1}{x_2}} \right)}^2} - 2\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + 4}}{{{x_1}{x_2} + \left( {{x_1} + {x_2}} \right) + 1}} = 4\\
\to \frac{{{{\left( {{x_1}{x_2}} \right)}^2} - 2{{\left( {{x_1} + {x_2}} \right)}^2} + 4{x_1}{x_2} + 4}}{{{x_1}{x_2} + \left( {{x_1} + {x_2}} \right) + 1}} = 4\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{\left( {{x_1} + {x_2}} \right)^2} + 4{x_1}{x_2} + 4 = 4{x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 4\\
\to {\left( {{x_1}{x_2}} \right)^2} - 2{\left( {{x_1} + {x_2}} \right)^2} - 4\left( {{x_1} + {x_2}} \right) = 0\\
\to {\left( {m - 2} \right)^2} - 2{\left( { - m} \right)^2} - 4\left( { - m} \right) = 0\\
\to {m^2} - 4m + 4 - 2{m^2} + 4m = 0\\
\to - {m^2} + 4 = 0\\
\to \left( {2 - m} \right)\left( {2 + m} \right) = 0\\
\to \left[ \begin{array}{l}
m = 2\\
m = - 2
\end{array} \right.
\end{array}\)
