Đáp án:
\(\begin{array}{l}
a)x = \dfrac{3}{4}\\
b)x = \dfrac{5}{2}\\
c)x = - \dfrac{{25}}{{18}}\\
d)x \in \emptyset \\
e)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4{x^2} + 4x + 1 - 9 + 12x - 4{x^2} = 4\\
\to 16x = 12\\
\to x = \dfrac{3}{4}\\
b)4{x^2} - 4x + 1 - 4{x^2} + 9 = 0\\
\to - 4x = - 10\\
\to x = \dfrac{5}{2}\\
c){x^2} - 6x + 9 + 9{x^2} + 12x + 4 - {x^2} + 16 = 9{x^2} - 12x + 4\\
\to 18x = - 25\\
\to x = - \dfrac{{25}}{{18}}\\
d){x^2} + 6x + 9 + 6 - 3x = 0\\
\to {x^2} + 3x + 15 = 0\\
\to {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{51}}{4} = 0\\
\to {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{51}}{4} = 0\left( {vô lý} \right)\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{51}}{4} > 0\forall x\\
\to x \in \emptyset \\
e)4\left( {4{x^2} + 4x + 1} \right) - 12{x^2} - 12x = 19\\
\to 16{x^2} + 16x + 4 - 12{x^2} - 12x - 19 = 0\\
\to 4{x^2} + 4x - 15 = 0\\
\to 4{x^2} + 4x + 1 - 16 = 0\\
\to {\left( {2x + 1} \right)^2} = 16\\
\to \left| {2x + 1} \right| = 4\\
\to \left[ \begin{array}{l}
2x + 1 = 4\\
2x + 1 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
