Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = 1\\
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = - 1\\
x = 2
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 7
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
e)x = 2\\
f)\left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}\\
x = - 3
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {x - 1} \right)\left( {{x^2} - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
{x^2} - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.\\
b)\left( {x + 1} \right)\left( {x - 1} \right) - \left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {x - 1 - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
x - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 2
\end{array} \right.\\
c)\left( {3x - 1} \right)\left( {2x - 5} \right) - \left( {3x - 1} \right)\left( {x + 2} \right) = 0\\
\to \left( {3x - 1} \right)\left( {2x - 5 - x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
3x - 2 = 0\\
x - 7 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 7
\end{array} \right.\\
d)\left( {x - 3} \right)\left( {3 - 4x} \right) + {\left( {x - 3} \right)^2} = 0\\
\to \left( {x - 3} \right)\left( {3 - 4x + x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
- 3x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
e)\left( {x - 2} \right)\left( {{x^2} + 1} \right) = 0\\
\to x - 2 = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
\to x = 2\\
f)x\left( {2{x^2} + 5x - 3} \right) = 0\\
\to x\left( {2x - 1} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}\\
x = - 3
\end{array} \right.
\end{array}\)
