giả thiết => \(\frac{M\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{N\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{32x-19}{\left(x+1\right)\left(x-2\right)}\)

=> M(x-2) + N(x+1) = 32x - 19

<=> M.x - 2.M + N.x + N = 32.x -19

=> (M+ N).x + (N - 2.M) = 32.x - 19

=> M+ N = 32 và -2M + N = -19 

=> M = 17, N = 15

vậy M.N = 17. 15 =...

a) \(x^3-4x^2+5x-2\)

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

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

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

=\(\left(x-1\right)\left(x^2-x-2x-2\right)\)

=\(\left(x-1\right)\left(\left(x\left(x-1\right)-2\left(x-1\right)\right)\right.\)

=\(\left(x-1\right)^2\left(x-2\right)\)

b) \(\) \(x^5+x+1\)

=\(x^5-x^2+x^2+x+1\)

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

=\(x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

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

c) \(x^3+5x^2+5x+1\)

=\(\left(x^3+1\right)+\left(5x^2+5x\right)\)

=\(\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)

=\(\left(x+1\right)\left(x^2+4x+1\right)\)

d) \(x^2\left(x^2+2y^2\right)-3y^4\)

=\(x^4+2x^2y^2-3y^4\)

=\(x^4-x^2y^2+3x^2y^2-3y^4\)

=\(x^2\left(x^2-y^2\right)+3y^2\left(x^2-y^2\right)\)

=\(\left(x^2-y^2\right)\left(x^2+3y^2\right)\)

=\(\left(x-y\right)\left(x+y\right)\left(x^2+3y^2\right)\)