Đáp án:
Để pt ${x^2} - 2x + m = 0$ có 2 nghiệm phân biệt x1; x2 thì:
$\begin{array}{l}
\Delta ' > 0\\
\Rightarrow 1 - m > 0\\
\Rightarrow m < 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m
\end{array} \right.\\
a)7{x_2} - 4{x_1} = 47\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
7{x_2} - 4{x_1} = 47
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = 2 - {x_2}\\
7{x_2} - 4\left( {2 - {x_2}} \right) = 47
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = 2 - {x_2}\\
11{x_2} = 55
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = 2 - {x_2}\\
{x_2} = 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = - 3\\
{x_2} = 5
\end{array} \right.\\
Do:{x_1}{x_2} = m\\
\Rightarrow \left( { - 3} \right).5 = m\\
\Rightarrow m = - 15\left( {tmdk} \right)\\
Vay\,m = - 15\\
b){x_1} = - 4{x_2}\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1} = - 4{x_2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = - \dfrac{2}{3}\\
{x_1} = \dfrac{8}{3}
\end{array} \right.\\
\Rightarrow m = \left( {\dfrac{{ - 2}}{3}} \right).\dfrac{8}{3} = \dfrac{{ - 16}}{9}\left( {tmdk} \right)\\
Vay\,m = \dfrac{{ - 16}}{9}
\end{array}$
(Giải thích:
$\begin{array}{l}
a = 1;b' = \frac{{ - 2}}{2} = - 1;c = m\\
Do:\Delta ' = {\left( { - b} \right)^2} - a.c\\
\Rightarrow \Delta ' = 1 - m\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{ - b}}{a} = \frac{{ - \left( { - 2} \right)}}{1} = 2\\
{x_1}.{x_2} = \frac{c}{a} = \frac{m}{1} = m
\end{array} \right.
\end{array}$)
