Đáp án:
c) 2) \(\left[ \begin{array}{l}
x = - 3\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = {x^3} + 27 - {x^3} - 7x\\
= 27 - 7x\\
DK:x \ne \pm 5\\
B = \dfrac{{x\left( {x + 5} \right) + \left( {4 - x} \right)\left( {x - 5} \right) - 13x + 25}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 5x - {x^2} + 9x - 20 - 13x + 25}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{x + 5}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{1}{{x - 5}}\\
b)4{x^2} - 8xy - 7x + 14y\\
= 4x\left( {x - 2y} \right) - 7\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {4x - 7} \right)\\
c)1){x^2} - 25 - {x^2} + 2x - 15 = 0\\
\to 2x = 40\\
\to x = 20\\
2){x^2}\left( {x + 3} \right) - 9\left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {{x^2} - 9} \right) = 0\\
\to \left( {x + 3} \right)\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x + 3 = 0\\
x - 3 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = - 3\\
x = 3
\end{array} \right.
\end{array}\)
