\(\text{b) }4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\\ =4\left(x^2-6x+9\right)-\left(4x^2-1\right)\\ =4x^2-24x+36-4x^2+1\\ =\left(4x^2-4x^2\right)-24x+\left(36+1\right)\\ =-24x+37\)
\(\text{c) }\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\\ \Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=6\\\Leftrightarrow x^2-8x+16-x^2+4=6\\ \Leftrightarrow\left(x^2-x^2\right)-8x+\left(16+4\right)=6\\ \Leftrightarrow-8x+20=6\\ \Leftrightarrow-8x=-14\\ \Leftrightarrow x=\dfrac{7}{4} \)
Vậy \(x=\dfrac{7}{4}\)
b) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\)
\(=4\left(x^2-6x+9\right)-\left(4x-1\right)\)
\(=4x^2-26x+36-4x+1\)
\(=4x^2-30x+37\)
