\(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|>=\left|2x+3+1-2x\right|=4\)
\(\dfrac{8}{3\left(x+1\right)^2+2}< =\dfrac{8}{2}=4\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+1=0\\\left(2x+3\right)\left(2x-1\right)< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-1\\-\dfrac{3}{2}< =x< =1\end{matrix}\right.\)
=>x=-1
\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3}\left(x+1\right)^2+2\left(1\right)\)
Ta có :
\(\left|2x+3\right|+\left|2x-1\right|\ge\left|2x+3+1-2x\right|=4\)
Dấu "=" xảy ra khi \(\left(2x+3\right)\left(1-2x\right)\ge0\Leftrightarrow-\dfrac{3}{2}\le x\le\dfrac{1}{2}\left(2\right)\)
\(\left(1\right)\Rightarrow\dfrac{8}{3}\left(x+1\right)^2+2=4\)
\(\Rightarrow\dfrac{8}{3}\left(x+1\right)^2=2\)
\(\Rightarrow\left(x+1\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=\dfrac{\sqrt{3}}{2}\\x+1=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}-1\\x=-\dfrac{\sqrt{3}}{2}-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}-2}{2}\left(tm\left(2\right)\right)\\x=-\dfrac{\sqrt{3}-2}{2}\left(kmt\left(2\right)\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{\sqrt{3}-2}{2}\) thỏa \(\left(1\right)\)
Kiểm tra thay \(x=\dfrac{\sqrt{3}-2}{2}\) vào \(\left(1\right)\) ta được :
\(\left|2.\dfrac{\sqrt{3}-2}{2}+3\right|+\left|2.\dfrac{\sqrt{3}-2}{2}-1\right|=\dfrac{8}{3}\left(\dfrac{\sqrt{3}-2}{2}+1\right)^2+2\)
\(\Rightarrow\left|\sqrt{3}-2+3\right|+\left|\sqrt{3}-2-1\right|=\dfrac{8}{3}.\dfrac{\left(\sqrt{3}\right)^2}{4}+2\)
\(\Rightarrow\sqrt{3}+1-\sqrt{3}+3=2+2=4\left(đúng\right)\)
