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

c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)

d: \(=\left(2x+5y\right)^3\)

a. (a² - b²)² - (a² + b²)²

= (a² - b² - a² - b²)(a² - b² + a² + b²)

= -2b² . 2a²

b. a⁶ - b⁶

<=> (a³)² - (b³)²

<=> (a³ - b³)(a³ + b³)

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

\(b,a^6-b^6=a^2\left(a^3-b^3\right)=a^2\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(c,-4x^2+9y^2=\left(3y-2x\right)\left(3y+2x\right)\\ d,\left(x+1\right)^3-\left(2-x\right)^3\\ =\left(x+1-2+x\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(2-x\right)+\left(2-x\right)^2\right]\\ =\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)\\ =\left(2x-1\right)\left(x^2-x+7\right)\)

\(e,8+\left(4x-3\right)^3\\ =\left(8+4x-3\right)\left[64-8\left(4x-3\right)+\left(4x-3\right)^2\right]\\ =\left(4x+5\right)\left(64-32x+24+16x^2-24x+9\right)\\ =\left(4x+5\right)\left(16x^2-56x+97\right)\)

\(g,81-\left(9-x^2\right)^2\\ =\left(9-9+x^2\right)\left(9+9-x^2\right)\\ =x^2\left(18-x^2\right)\left[=x^2\left(\sqrt{18}-x\right)\left(\sqrt{18}+x\right)\right]\)

Chỗ trong ngoặc nếu bạn chưa học căn thì ko cần ghi nha

a: Ta có: \(\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\)

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

\(=-4a^2b^2\)

b: \(a^6-b^6\)

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

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

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

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

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

\(=\left(2x-1\right)\left(x^2-x+7\right)\)

a) \(x^2-6x-y^2-4y+5=x^2-6x+9-y^2-4y-4\)

\(=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)=\left(x-3\right)^2-\left(y+2\right)^2\)

b) \(4a^2-12a-b^2+2b+8=4a^2-12a+9-b^2+2b-1\)

\(=\left(4a^2-12a+9\right)-\left(b^2-2b+1\right)=\left(2a-3\right)^2-\left(b-1\right)^2\)

x² - 6x - y² - 4y + 5

= ( x² - 6x + 9 ) - ( y² + 4y + 4 )

= ( x - 3 )² - ( y + 2 )²

4a² - 12a - b² + 2b + 8

= ( 4a² - 12a + 9 ) - ( b² - 2b + 1 )

= ( 2a - 3 )² - ( b - 1 )²

a) 

\(=x^2-6x+9-y^2-4y-4\) 

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

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

b) 

\(=4a^2-12a+9-b^2+2b-1\) 

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

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

a)  (2a - b)(b + 4a) + 2a(b - 3a)
= 2a(b + 4a) - b(b + 4a) + 2ab - 6a^2
= 2ab + 8a^2 - b^2 - 4ab + 2ab - 6a^2
= (8a^2 - 6a^2) + (2ab + 2ab - 4ab) - b^2
= 2a^2 - b^2
b) .(3a - 2b)(2a - 3b) - 6a(a - b)
= 3a(2a - 3b) - 2b(2a - 3b) - (6a^2 - 6ab)
= 6a^2 - 9ab - (4ab - 6b^2) - (6a^2 - 6ab)
= 6a^2 - 9ab - 4ab + 6b^2 - 6a^2 + 6ab
= 6b^2 + (6a^2 - 6a^2) + (6ab - 4ab - 9ab)
= 6b^2 - 7ab

c. 5b(2x - b) - (8b - x)(2x - b)
= 10bx - 5b^2 - 8b(2x - b) + x(2x - b)
= 10bx - 5b^2 - 16bx + 8b^2 + 2x^2 - bx
= (10bx - 16bx - bx) + 2x^2 + (8b^2 - 5b^2)
= -7bx + 2x^2 + 3b^2
d. 2x(a + 15x) + (x - 6a)(5a + 2x)
= 2ax + 30x^2 + x(5a + 2x) - 6a(5a + 2x)
= 2ax + 30x^2 + 5ax + 2x^2 - 30a^2 - 12ax
= (30x^2 + 2x^2) + (2ax + 5ax - 12ax) - 30a^2
= 32x^2 - 5ax - 30a^2

Chúc bạn hok tốt !!!

1) (a+2b+1)2

=a2+2a(2b+1)+(2b+1)2

=a2+4ab+2a+(2b)2+2.2b.1+12

=a2+4ab+2a+4b2+4b+1

2) (2a-b+3)2

=(2a)2 -2.2a(b-3)+(b-3)2

=4a2-4a(b-3)+b2-2b.3+32

=4a2-4ab+12a+b2 -6b+9

3) (2a-3b+1)2

=(2a)2-2.2a(3b-1)+(3b-1)2

=4a2-4a(3b-1)+(3b)2-2.3b.1+12

=4a2-4ab+4a+9b2-6b+1

a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)

\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)

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

b: \(\left(x+3\right)^2+4\left(x+3\right)+4\)

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

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

c: \(25+10\left(x+1\right)+\left(x+1\right)^2\)

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

\(=\left(x+6\right)\left(x+6\right)\)

a) \(2x^2+2b^2=x^2+b^2+x^2+b^2=x^2+2xb+b^2+x^2-2xb+b^2=\left(x+b\right)^2+\left(x-b\right)^2\)

       \(\dfrac{1}{4}a^2+2ab^2+4b^4\)

   =  \(\left(\dfrac{1}{2}a\right)^2+2\cdot a\cdot\dfrac{1}{2}+\left(2b\right)^2\)

   = \(\left(\dfrac{1}{2}a+2b\right)^2\)

   = \(\left(\dfrac{1}{2}a+2b\right)\cdot\left(\dfrac{1}{2}a+2b\right)\)