Question

Phân tích các đa thức sau thành nhân tử [ làm các bước luôn ạ, chứ đừng có ra kết quả ko thoi ạ ( - _ - ) ]

a/ 3x² - 6x

b/ x² - 2x + 1 - y²

c/ 9x³ - 9x²y - 4x + 4y

d/ x³ - 2x² - 8x

 

Answer 1

a) \(3x^2-6x\)

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

b) \(x^2-2x+1-y^2\)

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

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

Answer 2

c) \(9x^3-9x^2y-4x+4y\)

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

\(=9x^2\left(x-y\right)-4\left(x-y\right)\)

\(=\left(x-y\right)\left(9x^2-4\right)\)

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

Answer 3

a)\(3x^2-6x=3x\left(x-2\right)\)

b)\(x^2-2x+1-y^2=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left[\left(x-1\right)-y\right]\left[\left(x-1+y\right)\right]=\left(x-1-y\right)\left(x-1+y\right)\)

c)\(9x^3-9x^2y-4x+4y=\left(9x^3-9x^2y\right)-\left(4x-4y\right)=9x^2\left(x-y\right)-4\left(x-y\right)=\left(x-y\right)\left(9x^2-4\right)=\left(x-y\right)\left[\left(3x\right)^2-2^2\right]=\left(x-y\right)\left(3x-2\right)\left(3x+2\right)\)

d)\(x^3-2x^2-8x=x\left(x^2-2x-8\right)\)

Answer 4

c) \(9x^3-9x^2y-4x+4y\)

= \(9x^2\left(x-y\right)-4\left(x-y\right)\)

=\(\left(x-y\right)\left(9x^2-4\right)\)

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