4) \(2x^2+2x+1=\left(4x-1\right)\sqrt{x^2+1}\)
\(\Leftrightarrow\left[\left(4x-1\right)\sqrt{x^2+1}\right]^2=\left(2x^2+2x+1\right)^2\)
\(\Leftrightarrow\left(4x-1\right)^2.\left(x^2+1\right)=4x^4+4x^2+1+8x^3+4x^2+4x\)
\(\Leftrightarrow16x^4+16x^2-8x^3-8x+x^2+1=4x^4+8x^2+8x^3+4x+1\)
\(\Leftrightarrow16x^4+16x^2-8x^3-8x+x^2-4x^4-8x^2-8x^3-4x=-1+1\)
\(\Leftrightarrow16x^4-4x^4-8x^3-8x^3+16x^2+x^2-8x^2-8x-4x=0\)
\(\Leftrightarrow12x^4+9x^2-16x^3-12x=0\)
\(\Leftrightarrow x\left[3x\left(4x^2+3\right)-4\left(4x^2+3\right)\right]=0\)
\(\Leftrightarrow x\left(4x^2+3\right)\left(3x-4\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+3=0\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(lo\text{ại}\right)\\4x^2+3=0\left(v\text{ô}-l\text{ý}\right)\\x=\dfrac{4}{3}\left(nh\text{ậ}n\right)\end{matrix}\right.\)
S=\(\left\{\dfrac{4}{3}\right\}\)
