\(VP=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(VP=\dfrac{1}{2}\left[2x\left(x^2+1\right)+1\left(x^2+1\right)\right]\)
\(VP=\dfrac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(VT=\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}}\)
\(VT=\sqrt{x^2-\dfrac{1}{4}+\sqrt{\left(x+\dfrac{1}{2}\right)^2}}\)
\(VT=\sqrt{\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)+\left|x+\dfrac{1}{2}\right|}\)
Nhận thấy rằng: \(VT\ge0\Leftrightarrow VP\ge0\)
\(\Rightarrow\dfrac{1}{2}\left(2x+1\right)\left(x^2+1\right)\ge0\Leftrightarrow2x+1\ge0\Leftrightarrow x+\dfrac{1}{2}\ge0\)
Khi \(x+\dfrac{1}{2}\ge0\) thì: \(x+\dfrac{1}{2}=|x+\dfrac{1}{2}|\). Khi đó
\(VT=\sqrt{\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)+1\left(x+\dfrac{1}{2}\right)}\)
\(VT=\sqrt{\left(x-\dfrac{1}{2}+1\right)\left(x+\dfrac{1}{2}\right)}=\sqrt{\left(x+\dfrac{1}{2}\right)^2}=|x+\dfrac{1}{2}|=x+\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\left(2x+1\right)\left(x^2+1\right)=x+\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\left[2\left(x+\dfrac{1}{2}\right)\left(x^2+1\right)\right]=x+\dfrac{1}{2}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)\left(x^2+1\right)=x+\dfrac{1}{2}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)\left(x^2+1\right)-1\left(x+\dfrac{1}{2}\right)=0\)
\(\Rightarrow x^2\left(x+\dfrac{1}{2}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
