Chọn phát biểu đúng. Phương trình $a{x^2} + bx + c = 0\,\,(a \ne 0)$ có $a - b + c = 0$. Khi đó

$\left\{ \begin{array}{l}{x_1} + {x_2} =  - \dfrac{b}{a}\\{x_1}.{x_2} = \dfrac{c}{a}\end{array} \right.$

$\left\{ \begin{array}{l}{x_1} + {x_2} = \dfrac{b}{a}\\{x_1}.{x_2} = \dfrac{c}{a}\end{array} \right.$

$\left\{ \begin{array}{l}{x_1} + {x_2} =  - \dfrac{b}{a}\\{x_1}.{x_2} =  - \dfrac{c}{a}\end{array} \right.$

$\left\{ \begin{array}{l}{x_1} + {x_2} = \dfrac{b}{a}\\{x_1}.{x_2} =  - \dfrac{c}{a}\end{array} \right.$

${X^2} - PX + S = 0$

${X^2} - SX + P = 0$

$S{X^2} - X + P = 0$

${X^2} - 2SX + P = 0$

\({x_1} =  - 1;{x_2} = \dfrac{{1 - 2m}}{m}\)

\({x_1} = 1;{x_2} = \dfrac{{2m - 1}}{m}\)

\({x_1} = 1;{x_2} = \dfrac{{1 - 2m}}{m}\)

\({x_1} =  - 1;{x_2} = \dfrac{{2m - 1}}{m}\)

\({x_1} = 1;{x_2} =  - \dfrac{{26}}{5};B = \left( {x - 1} \right)\left( {x + \dfrac{{26}}{5}} \right)\)

\({x_1} = 1;{x_2} =  - \dfrac{{26}}{5};B = 5.\left( {x + 1} \right)\left( {x - \dfrac{{26}}{5}} \right)\)

\({x_1} = 1;{x_2} =  - \dfrac{{26}}{5};B = 5.\left( {x - 1} \right)\left( {x + \dfrac{{26}}{5}} \right)\)

\({x_1} = 1;{x_2} = \dfrac{{26}}{5};B = 5.\left( {x - 1} \right)\left( {x - \dfrac{{26}}{5}} \right)\)

\( - 6\)

\(16\)

\( - 16\)

\(6\)

\({x^2} - 4x - 3 = 0\)

\({x^2} + 3x - 4 = 0\)

\({x^2} - 4x + 3 = 0\)

\({x^2} + 4x + 3 = 0\)