Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\)
\(\Leftrightarrow4x^2-14x+10x-35-\left[-\left(4x+3\right)\right]^2=16\)
\(\Leftrightarrow4x^2-4x-35-\left(-1\right)^2\cdot\left(4x+3\right)^2-16=0\)
\(\Leftrightarrow4x^2-4x-35-\left(4x+3\right)^2-16=0\)
\(\Leftrightarrow4x^2-4x-35-\left(16x^2+24x+9\right)-16=0\)
\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)
\(\Leftrightarrow-12x^2-28x-60=0\)
\(\Leftrightarrow-12\left(x^2+\frac{28}{12}x+5\right)=0\)
\(\Leftrightarrow x^2+\frac{28}{12}x+5=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{7}{6}+\frac{49}{36}+\frac{131}{36}=0\)
\(\Leftrightarrow\left(x+\frac{7}{6}\right)^2+\frac{131}{36}=0\)(vô lý)
Vậy: \(S=\varnothing\)
