Đáp án:
e) \(\left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4{x^2} + 4x + 1 - 2x - 1 = 2\\
\to 4{x^2} + 2x - 2 = 0\\
\to 2{x^2} + x - 1 = 0\\
\to \left( {2x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - 1
\end{array} \right.\\
b)Đặt:{x^2} - 3x = t\\
Pt \to {t^2} + 5t + 6 = 0\\
\to {t^2} + 2t + 3t + 6 = 0\\
\to t\left( {t + 2} \right) + 3\left( {t + 2} \right) = 0\\
\to \left( {t + 2} \right)\left( {t + 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 2\\
t = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 3x = - 2\\
{x^2} - 3x = - 3\left( {vô nghiệm} \right)
\end{array} \right.\\
\to {x^2} - 3x + 2 = 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.\\
c)Đặt:{x^2} - x = t\\
Pt \to \left( {t - 1} \right)t - 2 = 0\\
\to {t^2} - t - 2 = 0\\
\to \left( {t - 2} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - x = 2\\
{x^2} - x = - 1\left( {vô nghiệm} \right)
\end{array} \right.\\
\to {x^2} - x - 2 = 0\\
\to \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
d)25 - 20x + 4{x^2} + 4x = 18\\
\to 4{x^2} - 16x + 7 = 0\\
\to \left( {2x - 7} \right)\left( {2x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
e)Đặt:{x^2} + 2x + 1 = t\left( {t \ge 0} \right)\\
Pt \to \left( {t + 2} \right)t = 3\\
\to {t^2} + 2t - 3 = 0\\
\to \left( {t + 3} \right)\left( {t - 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 3\left( l \right)\\
t = 1
\end{array} \right.\\
\to {x^2} + 2x + 1 = 1\\
\to {x^2} + 2x = 0\\
\to x\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
( câu f bạn xem lại đề nhé )
