Đáp án:
n) \(\left[ \begin{array}{l}
x = 4\\
x = - 1\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){\left( {x - 2} \right)^2} = {\left( {2x + 3} \right)^2}\\
\to \left| {x - 2} \right| = \left| {2x + 3} \right|\\
\to \left[ \begin{array}{l}
x - 2 = 2x + 3\\
x - 2 = - 2x - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 5\\
3x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{3}\\
x = - 5
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
b)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = \left( {x + 1} \right)\left( {2 - x} \right)\\
\to \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x + 1} \right)\left( {2 - x} \right) = 0\\
\to \left( {x + 1} \right)\left( {{x^2} - x + 1 - 2 + x} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
{x^2} - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 1
\end{array} \right.\\
c){x^2} - 4x - 5 = 0\\
\to {x^2} - 5x + x - 5 = 0\\
\to \left( {x - 5} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.\\
d){\left( {x + 3} \right)^3} = {\left( {x - 1} \right)^3}\\
\to x + 3 = x - 1\\
\to 3 = - 1\left( {voly} \right)\\
\to x \in \emptyset \\
e){\left( {2x + 1} \right)^2} - 2x - 3 = 0\\
\to 4{x^2} + 4x + 1 - 2x - 3 = 0\\
\to 4{x^2} + 2x - 2 = 0\\
\to \left( {2x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - 1
\end{array} \right.\\
g)Đặt:{x^2} - x = t\\
Pt \to \left( {t - 1} \right)t - 2 = 0\\
\to {t^2} - t - 2 = 0\\
\to \left( {t - 2} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - x = 2\\
{x^2} - x = - 1\left( {vô nghiệm} \right)
\end{array} \right.\\
\to {x^2} - x - 2 = 0\\
\to \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
h)\left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right.\left( {do:{x^2} - x + 1 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
k)\left( {x - 1} \right)\left( {x + 3} \right) = \left( {x - 1} \right)\left( {2x - 3} \right)\\
\to \left( {x - 1} \right)\left( {x + 3} \right) - \left( {x - 1} \right)\left( {2x - 3} \right) = 0\\
\to \left( {x - 1} \right)\left( {x + 3 - 2x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
- x + 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 6
\end{array} \right.\\
m){x^2} + {x^2} - 2 = 0\\
\to \left( {{x^2} - 1} \right)\left( {{x^2} + 2} \right) = 0\\
\to {x^2} - 1 = 0\left( {do:{x^2} + 2 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
n){x^2}\left( {x - 4} \right) - \left( {x - 4} \right) = 0\\
\to \left( {x - 4} \right)\left( {{x^2} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 1\\
x = 1
\end{array} \right.
\end{array}\)
