\(5,\\ a,M=\dfrac{\left(3x^2-2x-5\right)\left(2x-3\right)}{3x-5}=\dfrac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}=\left(x+1\right)\left(2x-3\right)\\ b,M=\dfrac{\left(2x^2+3x-2\right)\left(x^2-4x+4\right)}{x^2-4}=\dfrac{\left(x+2\right)\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\left(2x-1\right)\left(x-2\right)\)
\(6,\\ a,N=\dfrac{\left(x+1\right)\left(x^3+8\right)}{x^2-2x+4}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=\left(x+1\right)\left(x+2\right)\\ b,N=\dfrac{\left(2x^3-8x^2-6x+36\right)\left(x+3\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x-3\right)^2}{\left(x-3\right)\left(x+2\right)}=x-3\)