Đáp án:
15) \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{{\sqrt 3 - 1}}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(13)\Delta = 0 - 4.2.3 = - 24 < 0\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
16)\Delta = {\left( {\dfrac{7}{{12}}} \right)^2} - 4.\dfrac{1}{{12}}.\left( { - 19} \right)\\
= \dfrac{{961}}{{144}}\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - \dfrac{7}{{12}} + \sqrt {\dfrac{{961}}{{144}}} }}{{\dfrac{1}{6}}}\\
x = \dfrac{{ - \dfrac{7}{{12}} - \sqrt {\dfrac{{961}}{{144}}} }}{{\dfrac{1}{6}}}
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x = - 19
\end{array} \right.\\
19)\Delta = 9 - 4.5.0 = 9\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
14)\Delta = 29,8116\\
\to \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{{13}}{{10}}
\end{array} \right.\\
17)\Delta = 25 - 4.3.\left( { - 8} \right) = 121\\
\to \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = - 1
\end{array} \right.\\
20){x^2} - 4x + 1 = 0\\
\Delta ' = 4 - 1 = 3\\
\to \left[ \begin{array}{l}
x = 2 + \sqrt 3 \\
x = 2 - \sqrt 3
\end{array} \right.\\
15)\Delta ' = 3 - 4.\left( { - 1 + \sqrt 3 } \right) = 7 - 4\sqrt 3 \\
= 4 - 2.2\sqrt 3 + 3 = {\left( {2 - \sqrt 3 } \right)^2}\\
\to \left[ \begin{array}{l}
x = \dfrac{{\sqrt 3 + \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{4}\\
x = \dfrac{{\sqrt 3 - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{{\sqrt 3 - 1}}{2}
\end{array} \right.\\
21){x^2} + 3x + 2 - 70 = 0\\
\Delta = 9 - 4.\left( { - 68} \right) = 281\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 3 + \sqrt {281} }}{2}\\
x = \dfrac{{ - 3 - \sqrt {281} }}{2}
\end{array} \right.
\end{array}\)
