Question

\(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^{3-1}}\)=\(\frac{4}{x^2+x+1}\)

Answer 1

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)  ĐKXĐ : \(x^3-1\ne0\) <=> \(x^3\ne1\) <=> \(x\ne1\)

<=> \(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) 

=>  \(x^2+x+1+2x^2-5=4\left(x-1\right)\) 

<=> \(3x^2+x-4=4x-4\) 

<=> \(3x^2-3x=0\) 

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

<=> \(\begin{matrix}3x=0\\x-1=0\end{matrix}< =>\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\)

vậy  \(S=\left\{0\right\}\)

Answer 2

Bạn ghi lại đề đi bạn

Answer 3

Mày nhìn cái chóa j