(a - b)³ - (b - a)² + b(a - b)²

= (a - b)³ - ( a-b ) ² + b( a-b ) ²

= ( a-b )² . ( a - b - 1 + b )

= ( a² - 2ab + b² ) (a - 1 )

 vậy ....

Mình không chắc là đúng đâu nhé

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

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

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

Ta có:  x⁶ -y⁶= (x³)²  -(y³)²  = (x³  - y³)(x³ + y³)

\(x^6-y^6\)

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

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

hk 

tốt

1+ x² - y² -2 =

=x² -(y+1)²

= ( x+y+1)(x-y-1)

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

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

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

a) \(\left(a+b\right)^3+\left(a+b\right)^3\)

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

\(=2\left(a+b\right)\left[\left(a+b\right)^2\left(1-2+1\right)\right]\)

\(=2\left(a+b\right)\)

b)  \(9x^2+6xy+y^2\)

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

\(=\left(3x+y\right)\left(3x+y\right)\)

c)  \(4x^2-25\)

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

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

mình k ghi lại đề nha bạn

\(=\left(x-y\right)^2-16z^2\\ =\left(x-y-4z\right)\left(x-y+4z\right)\)

x²-6xy+9y²-16z²

=[x²-2.3xy+(3y)²]-16z²

=[x-3y]²-[4y]²

=[x-3y-3z][x-3y+3z]

Lời giải trước mk đánh nhầm nha

x²-6xy+9y²-16z²

=[x²-2.3xy+(3y)²]-16z²

=[x-3y]²-[4z]²

=[x-3y-4z][x-3y+4z]

a) 16(4x+5)² - 25(2x+2)²

^{\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)}

^{\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)}

^{\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)}

^{\(=\left(26x+30\right)\left(6x+10\right)\)}

\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-2y+1\right)\)

\(=\left(3x+2y+3\right)\left(-x-3y+5\right)\)

\(c,\left(x+1\right)^4-\left(x-1\right)^4\)

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

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

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

\(=\left(2x^2+2\right)2x.2\)

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

\(=8x\left(x^2+1\right)\)

\(d,16x^2-24xy+9y^2\)

\(=\left(4x\right)^2-2.4.3xy+\left(3y\right)^2\)

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

\(e,a^2-b^2+4bc-4c^2\)

\(=a^2-\left(b^2-4bc+4c^2\right)\)

\(=a^2-\left(b-2c\right)^2\)

\(=\left(a+b-2c\right)\left(a-b+2c\right)\)