Question

Phân tích \(x^2-2x-5=0\) thành nhân tử

Answer 1

`x^2-2x-5=0`

`<=>x^2-2x+1-6=0`

`<=>(x-1)^2-(\sqrt{6})^2=0`

`<=>(x-1-\sqrt{6})(x-1+\sqrt{6})=0`

`<=>` $\left[\begin{matrix} x=1+\sqrt{6}\\ x=1-\sqrt{6}\end{matrix}\right.$

Vậy `S={1+-\sqrt{6}}`

Answer 2

= x² + \(\sqrt{6}x\) - x - x - \(\sqrt{6}\) + 1 - \(\sqrt{6}x\) - 6 + \(\sqrt{6}\) 

= x( x +\(\sqrt{6}\) - 1) - ( x + \(\sqrt{6}\) - 1 ) - \(\sqrt{6}\)( x + \(\sqrt{6}\) - 1 ) 

= ( x+ \(\sqrt{6}\) - 1 ).( x - 1 - \(\sqrt{6}\) )