\(a,=x^2-9-x^2+6x-9=6x-18\\ b,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

a) \(A=\frac{2n+1+3n-5-4n+5}{n-3}=\frac{n+1}{n-3}\)

b) \(A=\frac{n+1}{n-3}=\frac{n-3+4}{n-3}=1+\frac{4}{n-3}\)

Để A đạt giá trị nguyên thì \(\frac{4}{n-3}\)đạt giá trị nguyên <=> \(n-3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

Tới đây lập bảng tìm n.

Lời giải:
$4n^3-4n^2-n+4=2n^2(2n-1)-n(2n-1)-(2n-1)+3$

$=(2n-1)(2n^2-n-1)+3$

Do đó để $4n^3-4n^2-n+4\vdots 2n-1$ thì:

$3\vdots 2n-1$
$\Rightarrow 2n-1\in\left\{1; -1;3;-3\right\}$

$\Rightarrow n\in \left\{1; 0; 2; -1\right\}$

Mà $n$ là số nguyên dương nên $n\in \left\{1;2\right\}$

a ) \(\left(a^6-3a^3+9\right)\left(a^3+3\right)=a^9+27\)

b ) Đặt \(a-y=t\) , ta có :

\(\left(t-x\right)^3-\left(t+x\right)^3\)

\(=\left(t-x-t-x\right)\left[\left(t-x\right)^2+\left(t-x\right)\left(t+x\right)+\left(t+x\right)^2\right]\)

\(=-2x\left[t^2-2tx+x^2+t^2-x^2+t^2+2tx+x^2\right]\)

\(=-2x\left[\left(t^2+t^2+t^2\right)+\left(x^2-x^2+x^2\right)+\left(2tx-2tx\right)\right]\)

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

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

\(=-2x\left(3a^2-6ay+3y^2+x^2\right)\)

c ) \(\left(4n^2-6mn+9m^2\right)\left(2n+3m\right)=8n^3+27m^3\)

d ) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)=125a^3-8b^3\)

a, ( a⁶ - 3a³ + 9 )(a³+ 3) = (a³)³ - 3³ = a⁹ - 27

b, ( a-x-y)³ - (a+x-y)³ = (a-x-y-a+x-y)(a-x-y+a+x-y)

= (-2y)(2a-2y) = -2y.2(a-y)

c, (4n²- 6mn + 9m²)(2n + 3m) = (2n)³ + (3m)³

= 8n³ + 27m³

d, (25a² + 10ab +4b²)( 5a - 2b ) = 125a³ - 8b³

 Cho A= ( 5m^2 - 8m^2 - 9m^2)( -n^3 + 4n^3)
Với giá trị nào m,n thì A ≥​ 0
A= ( 5m^2 - 8m^2 - 9m^2)( -n^3 + 4n^3)
A= -12m^2/3n^3
= -4m^2/n^3
do m^2>=0 với mọi m
nên A>=0
=> n<0 d0 -4<0

vậy A ≥​ 0 khi n<0 vầ m bất kì

b) Ta có: \(\dfrac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)

\(=\dfrac{4^{2m}-4^{2n}}{4^n+4^m}\)

\(=\dfrac{\left(4^m+4^n\right)\left(4^m-4^n\right)}{4^n+4^m}\)

\(=4^m-4^n\)

a) A=(3m+4n-5p)-(3m-4n-5p)

A= 3m+4n - 5p - 3m +  4n + 5p

A= 0