Đáp án:
$x = \dfrac{1}{2}$
Giải thích các bước giải:
$\sqrt{4x^2 + 4x+ 5} + \sqrt{8x^2 + 8x + 11} = 4 - 4x^2 - 4x$
$\Leftrightarrow \sqrt{4x^2 + 4x+ 5} - 2 + \sqrt{8x^2 + 8x + 11} - 3 + 4x^2 +4x +1 = 0$
$\Leftrightarrow \dfrac{(\sqrt{4x^2 + 4x+ 5} - 2)(\sqrt{4x^2 + 4x+ 5} + 2)}{\sqrt{4x^2 + 4x+ 5} +2} + \dfrac{(\sqrt{8x^2 + 8x + 11} - 3)(\sqrt{8x^2 + 8x + 11} + 3)}{\sqrt{8x^2 + 8x + 11} + 3} + (2x + 1)^2 = 0$
$\Leftrightarrow \dfrac{4x^2 + 4x + 1}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{8x^2 + 8x +2}{\sqrt{8x^2 + 8x + 11} +3} + (2x +1)^2 = 0$
$\Leftrightarrow \dfrac{(2x +1)^2}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{2(2x +1)^2}{\sqrt{8x^2 + 8x + 11} +3} + (2x +1)^2 = 0$
$\Leftrightarrow (2x +1)^2\left(\dfrac{1}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{2}{\sqrt{8x^2 + 8x + 11} +3} + 1\right) = 0$
$\Leftrightarrow \left[\begin{array}{l}2x + 1 = 0\\\dfrac{1}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{2}{\sqrt{8x^2 + 8x + 11} +3} + 1= 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{1}{2}\\\dfrac{1}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{2}{\sqrt{8x^2 + 8x + 11} +3} + 1= 0 \quad (*)\end{array}\right.$
Do $\dfrac{1}{\sqrt{4x^2 + 4x+ 5} + 2} + \dfrac{2}{\sqrt{8x^2 + 8x + 11} +3} + 1> 0, \, \forall x$
nên $(*)$ vô nghiệm
Vậy phương trình có nghiệm duy nhất $x = \dfrac{1}{2}$
