Question

[Hoàng Huy]

Phân tích đa thức thành nhân tử 

\(\dfrac{x^2}{4}\)-xy+y^2

x^2+x+\(\dfrac{1}{\text{4}}\)

x^2+2\(\sqrt{3}\)x+3

4x^2-1

Answer 1

[Nguyễn Huy Tú]

a, \(\dfrac{x^2}{4}-xy+y^2=\left(\dfrac{x}{2}\right)^2-xy+y^2=\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.y+y^2\)

\(=\left(\dfrac{x^2}{2}-y\right)^2\)

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d, \(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)

Answer 2

[Yeutoanhoc]

`x^2/4-2*x/2*y+y^2`

`=(x/2-y)^2`

`x^2+x+1/4`

`=x^2+2*x*1/2+(1/2)^2`

`=(x+1/2)^2`

`x^2+2sqrt3x+3`

`=x+2xsqrt3+sqrt3^2`

`=(x+sqrt3)^2`

`4x^2-1`

`=(2x)^2-1`

`=(2x-1)(2x+1)`