Đáp án:
a. \(\left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = 6\\
Pt \to {x^2} - 5x + 4 = 0\\
\to {x^2} - x - 4x + 4 = 0\\
\to x\left( {x - 1} \right) - 4\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.
\end{array}\)
b. Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to 25 - 4.\left( {m - 2} \right) > 0\\
\to 25 - 4m + 8 > 0\\
\to 33 > 4m\\
\to m < \dfrac{{33}}{4}\\
Có:\dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }} = \dfrac{3}{2}\\
\to \dfrac{{\sqrt {{x_1}} + \sqrt {{x_2}} }}{{\sqrt {{x_1}{x_2}} }} = \dfrac{3}{2}\\
\to \dfrac{{{x_1} + {x_2} + 2\sqrt {{x_1}{x_2}} }}{{{x_1}{x_2}}} = \dfrac{9}{4}\\
\to \dfrac{{5 + 2\sqrt {m - 2} }}{{m - 2}} = \dfrac{9}{4}\left( {m \ne 2} \right)\\
\to 20 + 8\sqrt {m - 2} = 9m - 18\\
\to 8\sqrt {m - 2} = 9m - 38\left( {DK:2 < m < \dfrac{{33}}{4}} \right)\\
\to 64\left( {m - 2} \right) = 81{m^2} - 684m + 1444\\
\to 81{m^2} - 748m + 1572 = 0\\
\to \left( {m - 6} \right)\left( {81m - 262} \right) = 0\\
\to \left[ \begin{array}{l}
m = 6\\
m = \dfrac{{262}}{{81}}
\end{array} \right.\left( {TM} \right)
\end{array}\)
