\(\sqrt{4x^2+4x+5}+\sqrt{8x^2+8x+11}=4-4x^2-4x\)
<=> \(\sqrt{\left(2x+1\right)^2+4}+\sqrt{2\left(2x+1\right)^2+9}=5-\left(2x+1\right)^2\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{\left(2x+1\right)^2+4}\ge2\\\sqrt{2\left(2x+1\right)^2+9}\ge3\end{matrix}\right.\)
=> VT \(\ge\) 5 mà VP \(\le\) 5
Mà VT = VP
=> 2x + 1 = 0
<=> x = \(\dfrac{-1}{2}\)