Đáp án:
a) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\Delta ' = 4 - 3 = 1\\
\to \left[ \begin{array}{l}
x = 2 + 1\\
x = 2 - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
b)\Delta = 1 - 4.5.2 = - 39 < 0
\end{array}\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
c)\Delta ' = 1 - \dfrac{1}{3}.\left( { - \dfrac{2}{3}} \right) = \dfrac{{11}}{9}\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {\dfrac{{11}}{9}} }}{{\dfrac{1}{3}}}\\
x = \dfrac{{1 - \sqrt {\dfrac{{11}}{9}} }}{{\dfrac{1}{3}}}
\end{array} \right. \to \left[ \begin{array}{l}
x = 3 + \sqrt {11} \\
x = 3 - \sqrt {11}
\end{array} \right.\\
d)\Delta = {\left( {1 - 2\sqrt 2 } \right)^2} - 4.2.\left( { - \sqrt 2 } \right) = 9 + 4\sqrt 2 \\
\to \left[ \begin{array}{l}
x = \dfrac{{1 - 2\sqrt 2 + \sqrt {9 + 4\sqrt 2 } }}{4}\\
x = \dfrac{{1 - 2\sqrt 2 - \sqrt {9 + 4\sqrt 2 } }}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 - 2\sqrt 2 + \sqrt {8 + 2.2\sqrt 2 + 1} }}{4}\\
x = \dfrac{{1 - 2\sqrt 2 - \sqrt {8 + 2.2\sqrt 2 + 1} }}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 - 2\sqrt 2 + \sqrt {{{\left( {2\sqrt 2 + 1} \right)}^2}} }}{4}\\
x = \dfrac{{1 - 2\sqrt 2 - \sqrt {{{\left( {2\sqrt 2 + 1} \right)}^2}} }}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 - 2\sqrt 2 + 2\sqrt 2 + 1}}{4}\\
x = \dfrac{{1 - 2\sqrt 2 - 2\sqrt 2 - 1}}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \sqrt 2
\end{array} \right.
\end{array}\)
