Đáp án:
Giải thích các bước giải:
Bài 4:
1. a. Đặt:
\(\begin{array}{l}
{x^2} + 7x = t\left( {t \ne 0} \right)\\
Pt \to {t^2} + 18t + 72 = 0\\
\to \left[ \begin{array}{l}
t = - 6\\
t = - 12
\end{array} \right. \to \left[ \begin{array}{l}
{x^2} + 7x = - 6\\
{x^2} + 7x = - 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 6\\
x = - 3\\
x = - 4
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
b.{x^4} - {x^3} - 2{x^3} + 2{x^2} + 2{x^2} - 2x - x + 1 = 0\\
\to {x^3}\left( {x - 1} \right) - 2{x^2}\left( {x - 1} \right) + 2x\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
{x^3} - 2{x^2} + 2x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
\left( {x - 1} \right)\left( {{x^2} - x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
{x^2} - x + 1 = 0\left( {voli} \right)
\end{array} \right. \to x = 1
\end{array}\)
