Ta có : \(N=\frac{2}{x-2}.\left(x^3-x^2-2x\right)=\frac{2x}{x-2}\left(x^2-2x+x-2\right)\)

             \(=\frac{2x}{x-2}\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)

a, \(5x-20y=5\left(x-4y\right)\)

b, \(x^3-9x=x\left(x^2-9\right)=x\left(x-3\right)\left(x+3\right)\)

c, \(x^2-2xy-y^2-16=-\left(x+y\right)^2-16=-\left(x+y-4\right)\left(x+y+4\right)\)

a) ĐKXĐ :  \(x\ne-3;x\ne2\)

b) \(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-9}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)

Để \(A\inℤ\Rightarrow x-3⋮x-2\)

=> \(x-2-1⋮x-2\)

Vì \(x-2⋮x-2\)

=> \(1⋮x-2\)

=> \(x-2\inƯ\left(1\right)\)

=> \(x-2\in\left\{1;-1\right\}\)

=> \(x\in\left\{3;1\right\}\)

Vậy \(x\in\left\{3;1\right\}\)là giá trị cần tìm

a + b , ĐK \(x\ne2;-3\)

\(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)

Bài làm 

a, \(2x^2\left(3x-5\right)=6x^3-10x^2\)

b, \(\left(12x^3y^2-18x^2y\right):2xy=\left(12x^3y^2:2xy\right)-\left(18x^2y:2xy\right)=6x^2y-9x\)

Bài làm 

\(\left(x^2-2\right)\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

=x²-x³-2+2x+x³+3³

⁼x²-x³-2+2x+x³+27

=x²+25+2x