a: \(M=6x+\dfrac{x^2+2x-x^2+2x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x^2-4\right)\left(x^2+4\right)-2x\left(x^2-4\right)}{4x}\)
\(=6+\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x^2-4\right)\left(x^2-2x+4\right)}{4x}\)
\(=6+x^2-2x+4=x^2-2x+10\)
b: Để \(M^2=M\) thì M=0 hoặc M=1
=>\(x\in\varnothing\)
c: Vì \(M=x^2-2x+10=\left(x-1\right)^2+9>0\)
nên \(M^2>M\forall x\)