Đáp án:
m. \(\left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
e.{\left( {3x - 1} \right)^2} = {\left( {x + 5} \right)^2}\\
\to \left| {3x - 1} \right| = \left| {x + 5} \right|\\
\to \left[ \begin{array}{l}
3x - 1 = x + 5\\
3x - 1 = - x - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 6\\
4x = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
f.{\left( {2x - 1} \right)^2} = {\left( {x - 3} \right)^2}\\
\to \left| {2x - 1} \right| = \left| {x - 3} \right|\\
\to \left[ \begin{array}{l}
2x - 1 = x - 3\\
2x - 1 = - x + 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
3x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = \dfrac{4}{3}
\end{array} \right.\\
g.4{x^2} - 4x + 1 - 4{x^2} + 1 = 0\\
\to - 4x + 2 = 0\\
\to x = \dfrac{1}{2}\\
g.{x^4} + 4{x^2} - {x^2} - 4 = 0\\
\to {x^4} + 3{x^2} - 4 = 0\\
\to {x^4} - {x^2} + 4{x^2} - 4 = 0\\
\to {x^2}\left( {{x^2} - 1} \right) + 4\left( {{x^2} - 1} \right) = 0\\
\to \left( {{x^2} - 1} \right)\left( {{x^2} + 4} \right) = 0\\
\to {x^2} - 1 = 0\\
\to \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
i.{x^4} - {x^3} + {x^2} - x = 0\\
\to {x^3}\left( {x - 1} \right) + x\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {{x^3} + x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
{x^3} + x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x\left( {{x^2} + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0
\end{array} \right.\\
k.4{x^2} - 25 - 4{x^2} - 14x + 10x + 35 = 0\\
\to - 4x + 10 = 0\\
\to x = \dfrac{5}{2}\\
l.\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - \left( {x - 2} \right)\left( {x - 12} \right) = 0\\
\to \left( {x - 2} \right)\left( {{x^2} + 2x + 4 - x + 12} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
{x^2} + x + 16 = 0\left( {vô nghiệm} \right)
\end{array} \right.\\
\to x = 2\\
m.2\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.
\end{array}\)
