Đáp án+Giải thích các bước giải:
$\dfrac{x+5}{x+2}+\dfrac{2}{x-3}=2(x\neq3;x\neq-2)$
$⇔\dfrac{(x+5)(x-3)+2(x+2)}{(x+2)(x-3)}=2$
$⇔\dfrac{x^2+2x-15+2x+4}{x^2-x-6}=2$
$⇔x^2+4x-11=2x^2-2x-12$
$⇔x^2-6x-1=0$
$⇔x^2-6x+9-10=0$
$⇔(x-3)^2-10=0$
$⇔(x-3+\sqrt{10})(x-3-\sqrt{10})=0$
$⇔\left[\begin{matrix}x-3+\sqrt{10}=0\\x-3-\sqrt{10}=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=3-\sqrt{10}(t/m)\\x=3+\sqrt{10}(t/m)\end{matrix}\right.$
Vậy `S={3-\sqrt{10};3+\sqrt{10}}`
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$x^2=2x-5x$
$⇔x^2=-3x$
$⇔x^2+3x=0$
$⇔x(x+3)=0$
$⇔\left[\begin{matrix}x=0\\x+3=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=0\\x=-3\end{matrix}\right.$
Vậy `S={0;-3}`
