Đáp án:
Câu 9:
\({x^2} - 6x + 8\)
Giải thích các bước giải:
\(\begin{array}{l}
C9:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
\left( {{x_1} + {x_2}} \right)\left( {{x_1} - {x_2}} \right) = - 12
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1} - {x_2} = - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1} = {x_2} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_2} - 2 + {x_2} = 6\\
{x_1} = {x_2} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2{x_2} = 8\\
{x_1} = {x_2} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_2} = 4\\
{x_1} = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1}{x_2} = 2.4 = 8
\end{array} \right.\\
\to Pt:{x^2} - Sx + P = {x^2} - 6x + 8\\
C10:\\
a.Thay:m = 3\\
Pt \to {x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array}\)
b. Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to {m^2} - 2m + 1 - 4m + 12 > 0\\
\to {m^2} - 6m + 13 > 0\\
\to {\left( {m - 3} \right)^2} + 4 > 0\left( {ld} \right)\forall m \in R\\
\to dpcm\\
Có:A = 1 - \left( {{x_1}^2 + {x_2}^2} \right)\\
= 1 - \left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right]\\
= 1 - \left[ {{{\left( {m - 1} \right)}^2} - 2\left( {m - 3} \right)} \right]\\
= 1 - \left( {{m^2} - 2m + 1 - 2m + 6} \right)\\
= 1 - {m^2} + 4m - 7\\
= - \left( {{m^2} - 4m + 6} \right)\\
= - {\left( {m - 2} \right)^2} - 2\\
Do:{\left( {m - 2} \right)^2} \ge 0\forall m \in R\\
\to - {\left( {m - 2} \right)^2} \le 0\\
\to - {\left( {m - 2} \right)^2} - 2 \le - 2\\
\to A \le - 2\\
\to MaxA = - 2\\
\Leftrightarrow m - 2 = 0\\
\Leftrightarrow m = 2
\end{array}\)
