Question

giúp mk nha khẩn cấp

Answer 1

a: =>(x+1)(x^2-x+1)-(x+1)(x+4)=0

=>(x+1)(x^2-2x-3)=0

=>(x+1)^2(x-3)=0

=>x=3 hoặc x=-1

b: =>x(x-6)-2(x-6)=0

=>(x-6)(x-2)=0

=>x=6 hoặc x=2

Answer 2

`a, (x+1)(x^2-x+1) - (x+1)(x+4) = 0`

`<=> (x+1)(x^2 - x +1 - x - 4) = 0`

`<=> (x+1)(x^2- 2x - 3) =0 

`<=> (x+1)(x-3)(x+1) = 0`

`<=>` \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

`b, x(x-6) - 2(x-6) = 0`

`<=> (x-2)(x-6) = 0`

`<=>` \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)