-Đặt \(t=\left(x^2-x+1\right)\)

\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

\(=t\left(t-4x\right)-x\left(t-4x\right)\)

\(=\left(t-4x\right)\left(t-x\right)\)

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

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

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

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

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

Bài 1:

$5x+10=5(x+2)$

Bài 2:

Tại $x=8$ thì $x^2+4x+4=(x+2)^2=(8+2)^2=10^2=100$

Bài 3:

$x^2-6x+9=x^2-2.3.x+3^2=(x-3)^2$

Bài 4:

Diện tích mảnh đất là:

$(x+5)(x-5)=24$

$\Leftrightarrow x^2-25=24$

$\Leftrightarrow x^2=49$

$\Rightarrow x=7$ (do $x>5$)

Chiều dài mảnh đất là: $x+5=7+5=12$ (m)

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

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

Bài 6:

c: \(9x^2+6x+1=\left(3x+1\right)^2\)

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

e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

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

\(x^3-x^2-x+1\)

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

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

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

a, \(x^2\) + 4\(x\) - y² + 4

= (\(x^2\) + 4\(x\) + 4) - y²

= (\(x\) + 2)² - y²

= (\(x\) + 2 - y)(\(x\) + 2 + y)

b, 2\(x^2\) - 18

= 2.(\(x^2\) -9)

= 2.(\(x\) -3).(\(x\) + 3)

c, \(x^3\) - \(x^2\) - \(x\) + 1

= (\(x^3\) + 1) - (\(x^2\) +  \(x\))

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

=(\(x\) + 1).(\(x^2\) - \(x\) + 1  - \(x\))

= (\(x\) + 1).(\(x\) - 1)²