Đáp án: \(\dfrac{x_1+x_2}{|x_1-x_2|}=\dfrac{a+b}{a-b}\)
Lời giải:
Theo Vi-et ta có:
$\left\{\begin{array}{l} {x_1} + {x_2} = \dfrac{{a + b}}{{ab}}\\{x_1}.{x_2} = \dfrac{1}{{ab}}\end{array} \right.$
\({({x_1} + {x_2})^2} = {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} = {x_1}^2 + {x_2}^2 - 2{x_1}{x_2} + 4{x_1}{x_2} = {({x_1} - {x_2})^2} + 4{x_1}{x_2}\)
\( \Rightarrow {({x_1} - {x_2})^2} = {({x_1} + {x_2})^2} - 4{x_1}{x_2} = \dfrac{{{{(a + b)}^2}}}{{{a^2}.{b^2}}} - \dfrac{4}{{ab}} = \dfrac{{{a^2} + {b^2} - 2ab}}{{{a^2}.{b^2}}} = {\left( {\dfrac{{a - b}}{{ab}}} \right)^2}\)
\( \Rightarrow \left| {{x_1} - {x_2}} \right| = \left| {\dfrac{{a - b}}{{ab}}} \right| = \dfrac{{a - b}}{{ab}}\)
\(\Rightarrow \dfrac{x_1+x_2}{|x_1-x_2|}=\dfrac{a+b}{a-b}\)
