Bài 1:
a) 3x²y - 9xy² + xy
= xy.(3x - 9y + 1)
b) 4x² + 4x + 1
= (2x)² + 2.2x.1 + (1)²
= (2x + 1)²
c) 4x² + 4x + 1 - y²
= (2x)² + 2.2x.1 + (1)² - (y)²
= (2x + 1)² - (y)²
= (2x + 1 - y)(2x + 1 + y)
d) 2x.(x + y) - 3x - 3y
= 2x.(x + y) - 3.(x + y)
= (x + y)(2x - 3)
Bài 2:
a) 2x.(x - 1) - x.(4 + 2x) = 20
⇔ 2x² - 2x - 4x - 2x² = 20
⇔ -6x = 20
⇔ x = -$\frac{10}{3}$
Vậy x = -$\frac{10}{3}$
b) x² - 2x + 1 = 0
⇔ (x - 1)² = 0
⇒ x - 1 = 0
⇔ x = 1
Vậy x = 1
c) x² - 7x + 12 = 0
⇔ x² - 3x - 4x + 12 = 0
⇔ x.(x - 3) - 4.(x - 3) = 0
⇔ (x - 3)(x - 4) = 0
⇔ \(\left[ \begin{array}{l}x - 3 = 0\\x - 4 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 3\\x = 4\end{array} \right.\)
Vậy x ∈ {3; 4}
