Đáp án:
e) \(\left[ \begin{array}{l}
x = - 4\\
x = - 10
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\Delta ' = 9 - 5.27 = - 126\\
\to x \in \emptyset \\
b)\Delta = {\left( {\sqrt 2 + \sqrt 8 } \right)^2} - 4.4 = 2\\
\to \left[ \begin{array}{l}
x = \dfrac{{\sqrt 2 + \sqrt 8 + \sqrt 2 }}{2}\\
x = \dfrac{{\sqrt 2 + \sqrt 8 - \sqrt 2 }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\sqrt 2 \\
x = \sqrt 2
\end{array} \right.\\
c)\Delta ' = 144 - 4.9 = 108\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 24 + \sqrt {108} }}{4} = \dfrac{{ - 12 + 3\sqrt 3 }}{2}\\
x = \dfrac{{ - 24 - \sqrt {108} }}{4} = \dfrac{{ - 12 - 3\sqrt 3 }}{2}
\end{array} \right.\\
d)x\left( {x + 3} \right)\left( {x + 1} \right)\left( {x + 2} \right) = 8\\
\to \left( {{x^2} + 3x} \right)\left( {{x^2} + 3x + 2} \right) = 8\\
Đặt:{x^2} + 3x = t\\
Pt \to t\left( {t + 2} \right) - 8 = 0\\
\to {t^2} + 2t - 8 = 0\\
\to \left[ \begin{array}{l}
t = 2\\
t = - 4
\end{array} \right. \to \left[ \begin{array}{l}
{x^2} + 3x = 2\\
{x^2} + 3x = - 4\left( {vô nghiệm} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 3 + \sqrt {17} }}{2}\\
x = \dfrac{{ - 3 - \sqrt {17} }}{2}
\end{array} \right.\\
e)\left( {x + 5} \right)\left( {x + 9} \right)\left( {x + 6} \right)\left( {x + 8} \right) = 40\\
\to \left( {{x^2} + 14x + 45} \right)\left( {{x^2} + 14x + 48} \right) = 40\\
Đặt:{x^2} + 14x + 45 = t\\
Pt \to t\left( {t + 3} \right) - 40 = 0\\
\to {t^2} + 3t - 40 = 0\\
\to \left[ \begin{array}{l}
t = 5\\
t = - 8
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 14x + 45 = 5\\
{x^2} + 14x + 45 = - 8\left( {vô nghiệm} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 4\\
x = - 10
\end{array} \right.
\end{array}\)
