Đáp án:
d. \(\left[ \begin{array}{l}
x = - 4\\
x = 9
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x + 15} \right)\left( {x + 15 + x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 15 = 0\\
2x + 14 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 15\\
x = - 7
\end{array} \right.\\
b.{x^2} - 14x + 49 = 64 - 32x + 4{x^2}\\
\to 3{x^2} - 18x + 15 = 0\\
\to 3{x^2} - 3x - 15x + 15 = 0\\
\to \left( {x - 1} \right)\left( {3x - 15} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
3x - 15 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\\
c.\left( {2x + 3} \right)\left( {2x + 3 - 7} \right) = 0\\
\to \left[ \begin{array}{l}
2x + 3 = 0\\
2x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \frac{3}{2}\\
x = 2
\end{array} \right.\\
d.\left( {x + 4} \right)\left( {x + 4 - 2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x + 4 = 0\\
- x + 9 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 4\\
x = 9
\end{array} \right.
\end{array}\)
