Đáp án:
Giải thích các bước giải:
a.
Đặt $\sqrt{x^2+2012}=t>0 ⇒2012=t^2-x^2$
Pt trở thành:
$x^4+t=t^2-x^2$
$⇔x^4-t^2+x^2+t=0$
$⇔(x^2-t)(x^2+t)+x^2+t=0$
$⇔(x^2+t)(x^2-t+1)=0$
$⇔x^2-t+1=0$ (do $t>0 ⇒x^2+t>0$)
$⇔x^2+1=t$
$⇔x^2+1=\sqrt{x^2+2012}$
$⇔x^4+2x^2+1=x^2+2012$
$⇔x^4+x^2-2011=0$
$⇒\left[ \begin{array}{l}x^2=\dfrac{-1+\sqrt{8045}}{2}\\x^2=\dfrac{-1-\sqrt{8045}}{2}<0(loại)\end{array} \right.$
$⇔x=±\sqrt{\dfrac{-1+\sqrt{8045}}{2}}$
b.
TXĐ: $D=R$
$⇔\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+21}-4+x^2+2x+1=0$
$⇔\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+21-16}{\sqrt{5x^2+10x+21}+4}+x^2+2x+1=0$
$⇔\dfrac{3(x^2+2x+1)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5(x^2+2x+1)}{\sqrt{5x^2+10x+21}+4}+x^2+2x+1=0$
$⇔(x^2+2x+1)\left( \dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+21}+4}+1\right)=0$
$⇔x^2+2x+1=0$ (ngoặc phía sau luôn dương)
$⇔x=-1$
