Đáp án:
2) \(\left[ \begin{array}{l}
m = \dfrac{{ - 1 + \sqrt 5 }}{4}\\
m = \dfrac{{ - 1 - \sqrt 5 }}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)a.DK:a \ge 0;a \ne 1\\
A = \left( {\dfrac{{a - 2\sqrt a + 1}}{{a + 1}}} \right):\left[ {\dfrac{1}{{1 + \sqrt a }} - \dfrac{{2\sqrt a }}{{\left( {1 + \sqrt a } \right)\left( {a + 1} \right)}}} \right]\\
= \dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{a + 1}}:\left[ {\dfrac{{a - 2\sqrt a + 1}}{{\left( {1 + \sqrt a } \right)\left( {a + 1} \right)}}} \right]\\
= \dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{a + 1}}.\dfrac{{\left( {1 + \sqrt a } \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= 1 + \sqrt a \\
b.Thay:a = 2021 - 2\sqrt {2020} \\
= 2020 - 2.\sqrt {2020} .1 + 1\\
= {\left( {\sqrt {2020} - 1} \right)^2}\\
\to A = 1 + \sqrt {{{\left( {\sqrt {2020} - 1} \right)}^2}} \\
= 1 + \sqrt {2020} - 1 = \sqrt {2020}
\end{array}\)
2) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to {m^2} + 2m + 1 - m > 0\\
\to {m^2} + m + 1 > 0\left( {ld} \right)\forall m \in R\\
Có:\dfrac{{2{x_1}^2 - {x_1} + 2{x_2}^2 - {x_2}}}{{{x_1}{x_2}}} = {x_1}{x_2} + \dfrac{3}{{{x_1}{x_2}}}\\
\to \dfrac{{2\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) - \left( {{x_1} + {x_2}} \right)}}{{{x_1}{x_2}}} = \dfrac{{3 + {{\left( {{x_1}{x_2}} \right)}^2}}}{{{x_1}{x_2}}}\\
\to 2{\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) = 3 + {\left( {{x_1} + {x_2}} \right)^2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) = 3\\
\to {\left( {2m + 2} \right)^2} - 4m - 2m - 2 = 3\\
\to 4{m^2} + 8m + 4 - 6m = 5\\
\to 4{m^2} + 2m - 1 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{ - 1 + \sqrt 5 }}{4}\\
m = \dfrac{{ - 1 - \sqrt 5 }}{4}
\end{array} \right.
\end{array}\)
