Bài 1:
a) (2x - 5)(2x + 5) - 4x(x - 3) - (12x + 7)
= 4x² - 25 - 4x² + 12x - 12x - 7
= -32
b) (2x³ - 5x² + 6x - 15) : (2x - 5)
= [(2x³ + 6x) - (5x² + 15)] : (2x - 5)
= [2x(x² + 3) - 5(x² + 3)] : (2x - 5)
= [(2x - 5)(x² + 3)] : (2x - 5)
= x² + 3
c) (x³ + 64y³) : (x + 4y)
= [(x + 4y)(x² - 4xy + 16y²)] : (x + 4y)
= x² - 4xy + 16y²
Bài 2:
Phân tích đa thức thành nhân tử
a) x³ - 4x² + 12x - 27
= (x³ - 27) - (4x² - 12x)
= (x - 3)(x² + 3x +9) - 4x( x - 3)
= (x - 3)(x² + 3x + 9 - 4x)
= (x -3)(x² - x + 9)
b) x³ - x + 3x²y + 3xy² + y³ - y
= (x³ + 3x²y + 3xy² + y³) - (x + y)
= (x + y)³ - (x + y)
= (x + y)[(x + y)² - 1)
= (x + y)(x + y - 1)(x + y + 1)
Tìm x
a) x² + 5x + 6 = 0
⇔ x² + 2x + 3x + 6 = 0
⇔ x(x + 2) + 3(x + 2) = 0
⇔ (x + 3)(x + 2) = 0
⇔ \(\left[ \begin{array}{l}x+3=0\\x+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-3\\x=-2\end{array} \right.\)
Vậy x = -3 hoặc x = -2
b) 2x² + 5x - 3 = 0
⇔ 2x² + 6x - x - 3 = 0
⇔ 2x(x + 3) - (x + 3) = 0
⇔ (2x - 1)(x + 3) = 0
⇔ \(\left[ \begin{array}{l}2x-1=0\\x+3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-3\end{array} \right.\)
Vậy x = $\frac{1}{2}$ hoặc x = -3
