Đáp án: ${X^2} - 10\sqrt 2 .X + 1 = 0$
Giải thích các bước giải:
$Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = - 1
\end{array} \right.$
Gọi pt cần tìm có dạng:
$\begin{array}{l}
{X^2} - S.X + P = 0\\
{X_1} = \left| {{{\left( {{x_1}} \right)}^3}} \right|;{X_2} = \left| {{{\left( {{x_2}} \right)}^3}} \right|\\
\left\{ \begin{array}{l}
S = {X_1} + {X_2} = \left| {x_1^3} \right| + \left| {x_2^3} \right|\\
P = {X_1}.{X_2} = \left| {x_1^3} \right|.\left| {x_2^3} \right|
\end{array} \right.\\
\Leftrightarrow S = {\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right)^3} - 3\left| {{x_1}{x_2}} \right|.\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right)\\
Xet:\left| {{x_1}} \right| + \left| {{x_2}} \right|\\
= \sqrt {{{\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right)}^2}} \\
= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2} + 2\left| {{x_1}{x_2}} \right|} \\
= \sqrt {{2^2} - 2.\left( { - 1} \right) + 2.\left| { - 1} \right|} \\
= \sqrt {4 + 2 + 2} = 2\sqrt 2 \\
\Leftrightarrow S = {\left( {2\sqrt 2 } \right)^3} - 3.\left| { - 1} \right|.2\sqrt 2 \\
= 16\sqrt 2 - 6\sqrt 2 \\
= 10\sqrt 2 \\
P = \left| {x_1^3} \right|.\left| {x_2^3} \right| = {\left( {\left| {{x_1}{x_2}} \right|} \right)^3} = 1\\
Vậy\,{X^2} - 10\sqrt 2 .X + 1 = 0
\end{array}$
