Question

\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)

Giúp mk với

Answer 1

`1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1)`

ĐK:`x ne 1`

`pt<=>(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x(x-1))/(x^3-1)`

`<=>x^2+x+1-3x^2=2x^2-2x`

`<=>4x^2-3x-1=0`

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

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

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

`x ne 1=>x-1 ne 0`

`<=>4x+1=0`

`<=>x=-1/4`

Vậy `S={-1/4}`

Answer 2

đk: x khác 1

\(\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2x}{x^2+x+1}=0\)

<=> \(\dfrac{x^2+x+1-3x^2-2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

<=> \(x^2+x+1-3x^2-2x^2+2x=0\)

<=> -4x² + 3x + 1 =0

<=> (1-x)(4x+1) = 0 

Mà x khác 1

<=> 4x + 1 = 0

<=> \(x=\dfrac{-1}{4}\)

Answer 3

(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x)*(x-1)/(x^3-1)

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

=>(-4x^2+3x+1)=0

mua máy casio về bấm giải ptr ra