Đáp án:
b. \(\left[ \begin{array}{l}
x = - \dfrac{1}{3}\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {{x^2} - 4} \right| + \left| x \right| = 2\\
\to {\left( {{x^2} - 4} \right)^2} + 2x\left( {{x^2} - 4} \right) + {x^2} = 4\\
\to {x^4} - 8{x^2} + 16 + 2{x^3} - 8x + {x^2} = 4\\
\to {x^4} + 2{x^3} - 7{x^2} - 8x + 12 = 0\\
\to {x^4} - {x^3} + 3{x^3} - 3{x^2} - 4{x^2} + 4x - 12x + 12 = 0\\
\to {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) - 4x\left( {x - 1} \right) - 12\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right) + \left( {{x^3} + 3{x^2} - 4x - 12} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
{x^3} + 3{x^2} - 4x - 12 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
{x^3} + 2{x^2} + {x^2} + 2x - 6x - 12 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
{x^2}\left( {x + 2} \right) + x\left( {x + 2} \right) - 6\left( {x + 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x + 2 = 0\\
{x^2} + x - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2\\
\left( {x - 2} \right)\left( {x + 3} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2\\
x = 2\\
x = - 3
\end{array} \right.\\
b.{\left( {x - 1} \right)^2} + 2\left( {x - 1} \right)\left( {2x + 3} \right) + {\left( {2x + 3} \right)^2} = 1\\
\to {x^2} - 2x + 1 + 4{x^2} + 2x - 6 + 4{x^2} + 12x + 9 = 1\\
\to 9{x^2} + 12x + 3 = 0\\
\to \left( {3x + 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{3}\\
x = - 1
\end{array} \right.
\end{array}\)
