Tìm x biết: x^2+4x-13=0

Đáp án:

 $x \in\{ -\sqrt{17}-2; \sqrt{17}-2\}$

Giải thích các bước giải:

 $x^2+4x-13=0$

$\Rightarrow x^2+ (\sqrt{17}+2-\sqrt{17}+2)x-(\sqrt{17}-2).(\sqrt{17}+2)=0$

$\Rightarrow x^2+ (\sqrt{17}+2)x-(\sqrt{17}-2)x-(\sqrt{17}-2).(\sqrt{17}+2)=0$

$\Rightarrow x.(x+ \sqrt{17}+2)-(\sqrt{17}-2)(x+\sqrt{17}+2)=0$

$\Rightarrow (x-\sqrt{17}+2)(x+\sqrt{17}+2)=0$

$\Rightarrow \left[ \begin{array}{l}x-\sqrt{17}+2=0\\x+\sqrt{17}+2=0\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=\sqrt{17}-2\\x=-\sqrt{17}-2\end{array} \right.$