a: \(=x\left(x^2+4x+4-z^2\right)\)

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

z còn câu b

a) $4x^2+4x+1$

$=(2x)^2+2\cdot2x\cdot1+1^2$

$=(2x+1)^2$

b) $x^2+6x-y^2+9$

$=(x^2+6x+9)-y^2$

$=(x^2+2\cdot x\cdot3+3^2)-y^2$

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

$=(x+3-y)(x+3+y)$

$\text{#}Toru$

a: \(4x^2+4x+1\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)

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

b: \(x^2+6x-y^2+9\)

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

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

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

Đề sai rồi bạn

1.A

2.C

3.B

4.C

a

c

b

c

a) \(x^4-4x^2-4x-1=\left(x^4-1\right)-4x\left(x+1\right)=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-4x\left(x+1\right)=\left(x+1\right)\left[\left(x^2+1\right)\left(x-1\right)-4x\right]=\left(x+1\right)\left(x^3-x^2+x-1-4x\right)=\left(x+1\right)\left(x^3-x^2-3x-1\right)\)

b) \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2=10xy^2\left(x^3-x^2-x+1\right)=10xy^2\left(x-1\right)^2\left(x+1\right)\)

a: \(x^4-4x^2-4x-1\)

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

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

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

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

b: \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2\)

\(=10xy^2\left(x^3-x^2-x+1\right)\)

\(=10xy^2\cdot\left[\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\right]\)

\(=10xy^2\cdot\left(x+1\right)\left(x-1\right)^2\)

\(a,=3xy\left(x-2y\right)\\ b,=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x+y+3\right)\left(x-y\right)\\ c,=x\left[\left(x+2\right)^2-y^2\right]=x\left(x+y+2\right)\left(x-y+2\right)\\ d,\Leftrightarrow x\left(x^2-4\right)=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

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

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

Câu 4:

D=(x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)x^2+8x+10\)

Câu 2:

b: \(4x^2-12x+9\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

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

Câu 1:

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

b: \(\left(3x+y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot y+3\cdot3x\cdot y^2+y^3\)

\(=27x^3+27x^2y+9xy^2+y^3\)

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

= (2x - 1)(2x + 1)

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

= (x - 2)(x + 2)(x + 4)

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³