\[\begin{array}{l}
a)(2x + 5)(2x - 7) - {( - 4x - 3)^2} = 16\\
\Leftrightarrow 4{x^2} - 14x + 10x - 35 - \left( {16{x^2} + 24x + 9} \right) = 16\\
\Leftrightarrow 4{x^2} - 4x - 35 - 16{x^2} - 24x - 9 = 16\\
\Leftrightarrow - 12{x^2} - 28x - 44 = 16\\
\Leftrightarrow 12{x^2} + 28x + 60 = 0\\
\Leftrightarrow 3{x^2} + 7x + 20 = 0\\
\Leftrightarrow 3\left( {{x^2} + \frac{7}{3}x + \frac{{49}}{{36}}} \right) + \frac{{191}}{{12}} = 0\\
\Leftrightarrow 3{\left( {x + \frac{7}{6}} \right)^2} + \frac{{191}}{{12}} = 0\\
do:3{\left( {x + \frac{7}{6}} \right)^2} \ge 0\\
= > 3{\left( {x + \frac{7}{6}} \right)^2} + \frac{{191}}{{12}} \ge \frac{{191}}{{12}} > 0
\end{array}\]

vậy pt vô nghiệm

\[\begin{array}{l}
b)(8{x^2} + 3)(8{x^2} - 3) - {(8{x^2} - 1)^2} = 22\\
\Leftrightarrow {\left( {8{x^2}} \right)^2} - {3^2} - \left( {64{x^4} - 16{x^2} + 1} \right) = 22\\
\Leftrightarrow 64{x^4} - 9 - 64{x^4} + 16{x^2} - 1 = 22\\
\Leftrightarrow 16{x^2} - 10 = 22\\
\Leftrightarrow 16{x^2} = 32\\
\Leftrightarrow {x^2} = 2\\
\Leftrightarrow x = \pm \sqrt 2 \\
vay:S = \{ \sqrt 2 ; - \sqrt 2 \} \\
c)2{(x - 1)^2} - {(2 - x)^2} - (x - 3)(x + 3) = 0\\
\Leftrightarrow 2\left( {{x^2} - 2x + 1} \right) - \left( {4 - 4x + {x^2}} \right) - \left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow 2{x^2} - 4x + 2 - 4 + 4x - {x^2} - {x^2} + 9 = 0\\
\Leftrightarrow 7 = 0
\end{array}\]

=>VÔ lý

vbayaj pt vô nghiệm