Đáp án:
d. \(x = \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.{(3x + 4)^2} - (3x - 1)(3x + 1) = 49\\
\to 9{x^2} + 24x + 16 - 9{x^2} + 1 = 49\\
\to 24x = 32\\
\to x = \dfrac{4}{3}\\
b.{x^2} - 4x + 4 = 9(x - 2)\\
\to {\left( {x - 2} \right)^2} - 9\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {x - 2 - 9} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
x - 11 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 11
\end{array} \right.\\
c.{x^2} - 25 = 3x - 15\\
\to \left( {x - 5} \right)\left( {x + 5} \right) - 3\left( {x - 5} \right) = 0\\
\to \left( {x - 5} \right)\left( {x + 5 - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x - 5 = 0\\
x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 2
\end{array} \right.\\
d.{(x - 1)^3} + 3{(x + 1)^2} = ({x^2} - 2x + 4)(x + 2)\\
\to {x^3} - 3{x^2} + 3x - 1 + 3\left( {{x^2} + 2x + 1} \right) = {x^3} + 2{x^2} - 2{x^2} - 4x + 4x + 8\\
\to - 3{x^2} + 3x - 1 + 3{x^2} + 6x + 3 = 8\\
\to 9x = 6\\
\to x = \dfrac{2}{3}
\end{array}\)
