Đáp án:
f) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)5{x^3}{y^3}\left( {3y - 4x + 6} \right)\\
c){x^2} + 6x + 9 + {x^2} - 4x + 4 = 2{x^2}\\
\to 2x = - 13\\
\to x = - \dfrac{{13}}{2}\\
d)\left( {3x - 5} \right)\left( {3x - 5 - x} \right) = 0\\
\to \left[ \begin{array}{l}
3x - 5 = 0\\
2x - 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = \dfrac{5}{2}
\end{array} \right.\\
e)x\left( {{x^2} - 25} \right) = 0\\
\to x\left( {x - 5} \right)\left( {x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 5\\
x = - 5
\end{array} \right.\\
f)6{x^2} - 18x - 6{x^2} - 9x = 0\\
\to - 27x = 0\\
\to x = 0
\end{array}\)
