Đáp án:
f. \(\left[ \begin{array}{l}
x = - 1\\
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
b.\left( {x + 1} \right)\left( {x + 4} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 5\\
\to \left( {{x^2} + 5x + 4} \right)\left( {{x^2} + 5x + 6} \right) = 5\\
Đặt:{x^2} + 5x + 4 = t\\
Pt \to t\left( {t + 2} \right) - 5 = 0\\
\to {t^2} + 2t - 5 = 0\\
\to {\left( {t + 1} \right)^2} - 6 = 0\\
\to \left[ \begin{array}{l}
t + 1 = \sqrt 6 \\
t + 1 = - \sqrt 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
t = - 1 + \sqrt 6 \\
t = - 1 - \sqrt 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 5x + 4 + 1 - \sqrt 6 = 0\\
{x^2} + 5x + 4 + 1 + \sqrt 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{5}{4} - \sqrt 6 = 0\\
{\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{5}{4} + \sqrt 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + \dfrac{5}{2} = \dfrac{5}{4} + \sqrt 6 \\
x + \dfrac{5}{2} = - \dfrac{5}{4} - \sqrt 6 \\
x + \dfrac{5}{2} = \dfrac{5}{4} - \sqrt 6 \\
x + \dfrac{5}{2} = - \dfrac{5}{4} + \sqrt 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{5}{4} + \sqrt 6 \\
x = - \dfrac{{15}}{4} - \sqrt 6 \\
x = - \dfrac{5}{4} - \sqrt 6 \\
x = - \dfrac{{15}}{4} + \sqrt 6
\end{array} \right.\\
d.DK:x \ne \left\{ {0;1} \right\}\\
\dfrac{{{x^2} - x + 1 - 2x\left( {x - 1} \right)}}{{x\left( {x - 1} \right)}} = 0\\
\to {x^2} - x + 1 - 2{x^2} + 2x = 0\\
\to - {x^2} + x + 1 = 0\\
Do: Δ= 1 + 4.1 = 5 > 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{2}\\
x = \dfrac{{1 - \sqrt 5 }}{2}
\end{array} \right.\\
e.\left[ \begin{array}{l}
{x^2} + 5x + 6 = 0\\
{x^2} - 4x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x + 2} \right)\left( {x + 3} \right) = 0\\
x = 2 + \sqrt 3 \\
x = 2 - \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = - 3\\
x = 2 + \sqrt 3 \\
x = 2 - \sqrt 3
\end{array} \right.\\
f.{x^3} + {x^2} - 4{x^2} - 4x + 3x + 3 = 0\\
\to {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) + 3\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
{x^2} - 4x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
\left( {x - 3} \right)\left( {x - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
