Đáp án:
b. m=2
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = 4\\
Pt \to {x^2} - 2x - 3 = 0\\
\to {x^2} + x - 3x - 3 = 0\\
\to x\left( {x + 1} \right) - 3\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
x - 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.
\end{array}\)
b. Để phương trình có 2 nghiệm phân biệt
⇒Δ>0
\(\begin{array}{l}
\to {m^2} - 4m + 4 + 12 > 0\\
\to {m^2} - 4m + 16 > 0\left( {ld} \right)\forall m \in R\\
Có:\sqrt {{x_1}^2 + 2020} - \sqrt {{x_2}^2 + 2020} = {x_1} + {x_2}\\
\to {x_1}^2 + 2020 + {x_2}^2 + 2020 - 2\sqrt {\left( {{x_1}^2 + 2020} \right)\left( {{x_2}^2 + 2020} \right)} = {\left( {{x_1} + {x_2}} \right)^2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 4040 - 2\sqrt {{{\left( {{x_1}{x_2}} \right)}^2} + 2020\left( {{x_1}^2 + {x_2}^2} \right) + {{2020}^2}} = {\left( {{x_1} + {x_2}} \right)^2}\\
\to - 2{x_1}{x_2} + 4040 - 2\sqrt {{{\left( {{x_1}{x_2}} \right)}^2} + 2020\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right] + {{2020}^2}} = 0\\
\to - 2.\left( { - 3} \right) + 4040 - 2\sqrt {9 + 2020\left( {{m^2} - 4m + 4 - 2.\left( { - 3} \right)} \right) + {{2020}^2}} = 0\\
\to 4046 - 2\sqrt {4080409 + 2020\left( {{m^2} - 4m + 10} \right)} = 0\\
\to \sqrt {4080409 + 2020\left( {{m^2} - 4m + 10} \right)} = 2023\\
\to 4080409 + 2020{m^2} - 8080m + 20200 = 4092529\\
\to 2020{m^2} - 8080m + 8080 = 0\\
\to m = 2
\end{array}\)
