Giải phương trình bậc 2 bằng cách sử dụng công thức nghiệm thu gọn
a, $\sqrt{3}$ $x^2$ – (1-$\sqrt{3}$)x – 1 = 0
b, $x^2$ – (2+$\sqrt{3}$)x + 2$\sqrt{3}$ = 0
c, $\sqrt{6}$ $x^2$ – 4$\sqrt{5}$ + 8 = 0
d, $x^2$ + 2(1+$\sqrt{5}$)x + 1 + 2$\sqrt{5}$ = 0
$\begin{array}{l} a)\sqrt 3 {x^2} – \left( {1 – \sqrt 3 } \right)x – 1 = 0\\ \Rightarrow \Delta = {\left( {1 – \sqrt 3 } \right)^2} – 4.\sqrt 3 .\left( { – 1} \right)\\ = 1 – 2\sqrt 3 + 3 + 4\sqrt 3 \\ = 1 + 2\sqrt 3 + 3\\ = {\left( {1 + \sqrt 3 } \right)^2}\\ \Rightarrow \sqrt \Delta = 1 + \sqrt 3 \\ \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{1 – \sqrt 3 + \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\ {x_2} = \dfrac{{1 – \sqrt 3 – \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = – 1 \end{array} \right.\\ b){x^2} – \left( {2 + \sqrt 3 } \right)x + 2\sqrt 3 = 0\\ \Rightarrow \Delta = {\left( {2 + \sqrt 3 } \right)^2} – 4.2\sqrt 3 \\ = 4 + 4\sqrt 3 + 3 – 8\sqrt 3 \\ = 4 – 4\sqrt 3 + 3\\ = {\left( {2 – \sqrt 3 } \right)^2}\\ \Rightarrow \sqrt \Delta = 2 – \sqrt 3 \\ \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{2 + \sqrt 3 + \left( {2 – \sqrt 3 } \right)}}{2} = 2\\ {x_1} = \dfrac{{2 + \sqrt 3 – \left( {2 – \sqrt 3 } \right)}}{2} = \sqrt 3 \end{array} \right.\\ c)\sqrt 6 {x^2} – 4\sqrt 5 + 8 = 0\\ \Rightarrow {x^2} = \dfrac{{4\sqrt 5 – 8}}{{\sqrt 6 }} = \dfrac{{2\sqrt {30} – 4\sqrt 6 }}{3}\\ \Rightarrow x = \pm 0,62\\ d){x^2} + 2\left( {1 + \sqrt 5 } \right)x + 1 + 2\sqrt 5 = 0\\ \Rightarrow \Delta ‘ = {\left( {1 + \sqrt 5 } \right)^2} – 1 – 2\sqrt 5 \\ = 1 + 2\sqrt 5 + 5 – 1 – 2\sqrt 5 \\ = 5\\ \Rightarrow \sqrt {\Delta ‘} = \sqrt 5 \\ \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{1 + \sqrt 5 + \sqrt 5 }}{1} = 2\sqrt 5 + 1\\ {x_2} = 1 \end{array} \right. \end{array}$
Đáp án:
$\begin{array}{l}
a)\sqrt 3 {x^2} – \left( {1 – \sqrt 3 } \right)x – 1 = 0\\
\Rightarrow \Delta = {\left( {1 – \sqrt 3 } \right)^2} – 4.\sqrt 3 .\left( { – 1} \right)\\
= 1 – 2\sqrt 3 + 3 + 4\sqrt 3 \\
= 1 + 2\sqrt 3 + 3\\
= {\left( {1 + \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt \Delta = 1 + \sqrt 3 \\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{1 – \sqrt 3 + \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\
{x_2} = \dfrac{{1 – \sqrt 3 – \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = – 1
\end{array} \right.\\
b){x^2} – \left( {2 + \sqrt 3 } \right)x + 2\sqrt 3 = 0\\
\Rightarrow \Delta = {\left( {2 + \sqrt 3 } \right)^2} – 4.2\sqrt 3 \\
= 4 + 4\sqrt 3 + 3 – 8\sqrt 3 \\
= 4 – 4\sqrt 3 + 3\\
= {\left( {2 – \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt \Delta = 2 – \sqrt 3 \\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{2 + \sqrt 3 + \left( {2 – \sqrt 3 } \right)}}{2} = 2\\
{x_1} = \dfrac{{2 + \sqrt 3 – \left( {2 – \sqrt 3 } \right)}}{2} = \sqrt 3
\end{array} \right.\\
c)\sqrt 6 {x^2} – 4\sqrt 5 + 8 = 0\\
\Rightarrow {x^2} = \dfrac{{4\sqrt 5 – 8}}{{\sqrt 6 }} = \dfrac{{2\sqrt {30} – 4\sqrt 6 }}{3}\\
\Rightarrow x = \pm 0,62\\
d){x^2} + 2\left( {1 + \sqrt 5 } \right)x + 1 + 2\sqrt 5 = 0\\
\Rightarrow \Delta ‘ = {\left( {1 + \sqrt 5 } \right)^2} – 1 – 2\sqrt 5 \\
= 1 + 2\sqrt 5 + 5 – 1 – 2\sqrt 5 \\
= 5\\
\Rightarrow \sqrt {\Delta ‘} = \sqrt 5 \\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{1 + \sqrt 5 + \sqrt 5 }}{1} = 2\sqrt 5 + 1\\
{x_2} = 1
\end{array} \right.
\end{array}$