Đáp án:
\(\left[ \begin{array}{l}
x = - 2 + \sqrt { - 1 + 2\sqrt {11} } \\
x = - 2 - \sqrt { - 1 + 2\sqrt {11} }
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{{{({x^2} - 4)}^2} + 2{{(x - 2)}^2} = 43}\\
{ \to {{\left( {x - 2} \right)}^2}{{\left( {x + 2} \right)}^2} + 2{{\left( {x - 2} \right)}^2} = 43}\\
{ \to {{\left( {x - 2} \right)}^2}\left[ {{{\left( {x + 2} \right)}^2} + 2} \right] = 43}\\
{ \to {{\left( {x + 2} \right)}^2}\left( {{x^2} + 4x + 4 + 2} \right) = 43}\\
{ \to {{\left( {x + 2} \right)}^2}\left( {{x^2} + 4x + 6} \right) = 43}\\
{ \to \left( {{x^2} + 4x + 4} \right)\left( {{x^2} + 4x + 6} \right) = 43\left( 1 \right)}\\
{Đặt:{x^2} + 4x + 4 = t}\\
{\left( 1 \right) \to t\left( {t + 2} \right) - 43 = 0}\\
{ \to {t^2} + 2t - 43 = 0}\\
{ \to {t^2} + 2t + 1 = 44 \to {{\left( {t + 1} \right)}^2} = 44}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{t + 1 = 2\sqrt {11} }\\
{t + 1 = {\rm{\;}} - 2\sqrt {11} {\rm{\;}}}
\end{array} \to \left[ \begin{array}{l}
t = - 1 + 2\sqrt {11} \\
t = - 1 - 2\sqrt {11}
\end{array} \right.} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{{x^2} + 4x + 4 = {\rm{\;}} - 1 + 2\sqrt {11} }\\
{{x^2} + 4x + 4 = {\rm{\;}} - 1 - 2\sqrt {11} {\rm{\;}}}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{{{\left( {x + 2} \right)}^2} = - 1 + 2\sqrt {11} }\\
{{{\left( {x + 2} \right)}^2} = - 1 - 2\sqrt {11} \left( {voly:do:{{\left( {x + 2} \right)}^2} \ge 0\forall x \in R} \right)}
\end{array}} \right.}\\
{ \to \left[ \begin{array}{l}
x + 2 = \sqrt { - 1 + 2\sqrt {11} } \\
x + 2 = - \sqrt { - 1 + 2\sqrt {11} }
\end{array} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = - 2 + \sqrt { - 1 + 2\sqrt {11} } }\\
{x = - 2 - \sqrt { - 1 + 2\sqrt {11} } {\rm{\;}}}
\end{array}} \right.}\\
{}\\
{KL:\left[ {\begin{array}{*{20}{l}}
{x = {\rm{\;}} - 2 + \sqrt { - 1 + 2\sqrt {11} } }\\
{x = {\rm{\;}} - 2 - \sqrt { - 1 + 2\sqrt {11} } {\rm{\;}}}
\end{array}} \right.}
\end{array}\)
