Question

phân tích thành nhân tử

a)4 x^2-1/9

b)3x^2-6x+3

c)ax-4x+2a-8

d) x^2-2xy+y^2-16

Answer 1

a)\(4x^2-\dfrac{1}{9}=\left(2x\right)^2-\left(\dfrac{1}{3}\right)^2=\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)\)

b)\(3x^2-6x+3=3\left(x^2-2x+1\right)=3\left(x-1\right)^2\)

c)\(ax-4x+2a-8=\left(ax-4x\right)+\left(2a-8\right)=x\left(a-4\right)+2\left(a-4\right)=\left(a-4\right)\left(x+2\right)\)

d)\(x^2-2xy+y^2-16=\left(x^2-2xy+y^2\right)-16=\left(x-y\right)^2-16=\left(x-y\right)^2-4^2=\left(x-y-4\right)\left(x-y+4\right)\)

Answer 2

a) 4x² −\(\dfrac{1}{9}\)=(2x )²−(\(\dfrac{1}{3}\))²=(2x−\(\dfrac{1}{3}\))(2x+\(\dfrac{1}{3}\))

b)3x²−6x+3=3(x²−2x+1)