Question

phân tích đa thức thành nhân tử

27x³+8

b,x²-2xy+y²-3x+3y

c. x³-3x+2

Answer 1

a/ 27x³ + 8

= (3x)³ + 2³

= (3x+2).[(3x)² - 3x.2 + 2²]

= (3x+2).(9x² - 6x + 4)

b/ x² - 2xy + y² - 3x + 3y

= (x-y)² - 3.(x-y)

= (x-y).(x-y-3)

Answer 2

a) 27x³+8

= (3x)³+2³

=(3x+2)(9x²-6x+4)

b,x²-2xy+y²-3x+3y

=(x²-2xy+y²)-(3x-3y)

= (x-y)²-3(x-y)

= (x-y)(x-y-3)

c. x³-3x+2

= x³+0x²-3x+2

=x³+2x²-2x²-4x+x+2

=(x³+2x²)-(2x²+4x)+(x+2)

= x²(x+2)-2x(x+2)+(x+2)

=(x+2)(x²-2x+1)

=(x+2)(x-1)²

Answer 3

c)x³-3x+2

=(x³-x² )+(x²-x)-(2x-2)

=x²(x-1)+x(x-1)-2(x-1)

=(x²+x-2)(x-1)

=(x²-x+2x-2)(x-1)

=(x+2)(x-1)²

Answer 4

a) \(27x^3+8\)

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

= \(\left(3x+2\right)\left[\left(3x\right)^2+3x\cdot2+2^2\right]\)

= \(\left(3x+2\right)\left(9x^2+6x+4\right)\)

b) \(x^2-2xy+y^2-3x+3y\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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