Đáp án:
e) \(\left[ \begin{array}{l}
x = 2 + \sqrt {33} \\
x = 2 - \sqrt {33}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} = 12\\
\to \left[ \begin{array}{l}
x = 2\sqrt 3 \\
x = - 2\sqrt 3
\end{array} \right.\\
b)Do:\dfrac{1}{2}{x^2} + 2 > 0\forall x\\
\to \dfrac{1}{2}{x^2} + 2 = 0\left( {vô nghiệm} \right)\\
c) - \dfrac{9}{{10}}{x^2} + \dfrac{{36}}{{10}}x = 0\\
\to - 9{x^2} + 36x = 0\\
\to - 9x\left( {x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
d)\Delta ' = 4 + 1 = 5\\
\to \left[ \begin{array}{l}
x = - 2 + \sqrt 5 \\
x = - 2 - \sqrt 5
\end{array} \right.\\
e)\Delta ' = 4 + 32 = 33\\
\to \left[ \begin{array}{l}
x = 2 + \sqrt {33} \\
x = 2 - \sqrt {33}
\end{array} \right.\\
f)2{x^2} - \left( {4\sqrt 2 - \sqrt 3 } \right)x + 2\sqrt 6 = 0\\
\Delta = 35 - 8\sqrt 6 - 8.2\sqrt 6 \\
= 35 - 24\sqrt 6 \\
\to \left[ \begin{array}{l}
x = \dfrac{{4\sqrt 2 - \sqrt 3 + \sqrt {35 - 24\sqrt 6 } }}{4}\\
x = \dfrac{{4\sqrt 2 - \sqrt 3 - \sqrt {35 - 24\sqrt 6 } }}{4}
\end{array} \right.
\end{array}\)
