a) x⁴ + 2x³ + 2x² + 2x + 1

=[ (x²)² + 2.x².x + x² ] + (x² + 2.x.1 + 1²)

=(x² + x)² + (x + 1)²

=[x(x + 1)]² + (x + 1)²

=x²(x + 1)² + (x + 1)²

=(x² + 1)(x + 1)²

b) x² - 2x - 4y² - 4y

= [x² - (2y)²] - 2(x - 2y)

= (x - 2y)(x + 2y) - 2(x - 2y)

= (x + 2y - 2)(x - 2y)

x² - 4y² + x + 2y

= ( x² - 4y² ) + ( x + 2y )

= ( x - 2y ) ( x + 2y ) + ( x + 2y )

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

Bài 2: 

Sửa đề:  \(x^3-3x^2-10x=0\)

\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)

1: =(16x^2-8x+1)-y^2

=(4x-1)^2-y^2

=(4x-1-y)(4x-1+y)

2: =(x^2-2xy+y^2)-z^2

=(x-y)^2-z^2

=(x-y-z)(x-y+z)

3: =(x^2+4xy+4y^2)-16

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

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

4: =(x^2-4xy+4y^2)-16

=(x-2y)^2-4^2

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

\(\left(x+2y-4\right)\left(x+2y+4\right)\)

(x+y)^2 - 5y^2

x^2- 4y^2 + 4xy

= x^2 + 4xy - 4y^2

=x^2 + 2x2y - (2y)^2

= ( x - 2y )^2

 x² - 4y² + 4xy = x² + 4xy - 4y² = - (x² - 4xy + 4y²) = - ((x² - 2x.2y + (2y)²)

= - ( x - 2y)²

(x-2)*(x-1)

(x-2)(x-1)

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

\(x^2-25-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-25\)

\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)

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

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