a) ( 4n² - 6mn + 9m² ) . ( 2n + 3m ) = 

=  ( 2n + 3m ) . ( 4n² - 6mn + 9m² )

=          8n³ + 27m³ 

b) ( 7 + 2b ) . ( 4b² - 14b + 49 ) = 

=  ( 2b + 7 ) . ( 4b² - 14b + 49 )

=            8b³ + 343 

MÌnh nghĩ là chỗ - 4b phải viết là -14b

c) ( 25a² + 10ab + 4b² ) . ( 5a - 2b ) = 

= ( 5a - 2b ) . ( 25a² + 10ab + 4b² )

=            125a³ - 8b³ 

HOk tốt!!!!!!!!!!!!

a) \(\left(4n^2-6nm+9m^2\right)\left(2n+3m\right)\)

\(=\left(2n+3m\right)\left[\left(2n\right)^2-2n.3m+\left(3m\right)^2\right]\)

\(=\left(2n\right)^3+\left(3m\right)^3\)

\(=8n^3+27m^3\)

b) Sửa đề \(\left(7+2b\right)\left(4b^2-14b+49\right)\)

\(=\left(7+2b\right)\left[\left(2b\right)^2-2b.7+7^2\right]\)

\(=7^3+\left(2b\right)^3\)

\(=343+8b^3\)

c) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)\)

\(=\left(5a-2b\right)\left[\left(5a\right)^2+5a.2b+\left(2b\right)^2\right]\)

\(=\left(5a\right)^3-\left(2b\right)^3\)

\(=125a^3-8b^3\)

d) \(\left(x^2+x+2\right)\left(x^2-x-2\right)\)

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

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

Bài 1: 

a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)

\(=k^3-64-k^3-128\)

=-192

b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)

\(=8m^3+27n^3-27m^3+8n^3\)

\(=-19m^3+35n^3\)

Bài 4: 

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x=9\)

hay x=1

b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x=7\)

hay \(x=\dfrac{7}{2}\)

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

1: 2a+2b=2(a+b)

2: 2a+4b+6c

=2*a+2*2b+2*3c

=2(a+2b+3c)

3: \(-7a-14ab-21b=-7\left(a+2ab+3b\right)\)

4: \(2ax-2ay+2a=2a\left(x-y+1\right)\)

5: \(=3a\cdot ax-3a\cdot2ay+3a\cdot4=3a\left(ax-2ay+4\right)\)

6: \(=2\cdot2ax-2\cdot ay-2\cdot1=2\cdot\left(2ax-ay-1\right)\)

7: =a^2-(2b)^2

=(a-2b)(a+2b)

8: =(5a)^2-1^2

=(5a-1)(5a+1)

9: =9(16a^2-9)

=9(4a-3)(4a+3)

a) Rút gọn được A = ( k 3  – 64) – (128 +  k 3 ) = -192.

b) Rút gọn được B = -19 m 3  + 35 n 3 .

a) (a - 2b)x(a + 2b)
b) x²-(y-3)²
 => (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))² - (5(x + y))²
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)²
f) (5x - 2y)²
h) (x - 5)(x² + 5x + 25)

k) (x + 5)³

A.

$a^2+4b^2+9c^2=2ab+6bc+3ac$

$\Leftrightarrow a^2+4b^2+9c^2-2ab-6bc-3ac=0$

$\Leftrightarrow 2a^2+8b^2+18c^2-4ab-12bc-6ac=0$

$\Leftrightarrow (a^2+4b^2-4ab)+(a^2+9c^2-6ac)+(4b^2+9c^2-12bc)=0$

$\Leftrightarrow (a-2b)^2+(a-3c)^2+(2b-3c)^2=0$

$\Rightarrow a-2b=a-3c=2b-3c=0$

$\Rightarrow A=(0+1)^{2022}+(0-1)^{2023}+(0+1)^{2024}=1+(-1)+1=1$

B.

$x^2+2xy+6x+6y+2y^2+8=0$

$\Leftrightarrow (x^2+2xy+y^2)+y^2+6x+6y+8=0$

$\Leftrightarrow (x+y)^2+6(x+y)+9+y^2-1=0$

$\Leftrightarrow (x+y+3)^2=1-y^2\leq 1$ (do $y^2\geq 0$ với mọi $y$)

$\Rightarrow -1\leq x+y+3\leq 1$

$\Rightarrow -4\leq x+y\leq -2$

$\Rightarrow 2020\leq x+y+2024\leq 2022$

$\Rightarrow A_{\min}=2020; A_{\max}=2022$