\(ĐKXĐ:x\ne1\\ \dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\\ \Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Leftrightarrow\dfrac{-4x^2+3x+1}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Rightarrow4x^2-3x-1=0\)
`<=>4x^2 -4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>(4x+1)(x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy `S={-1/4}`
đkxđ : x khác 1
ta có pttđ :
\(\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{x^3-1}-\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
<=> \(\dfrac{x^2+x+1}{...}-\dfrac{3x^2}{...}-\dfrac{2x^2-2x}{..}=0\)
<=> \(x^2+x+1-3x^2-2x^2+2x=0\)
<=> \(-4x^2+3x+1=0\)
\(\Delta=3^2-4.-4.1=25;\sqrt{\Delta}=5>0\)
=> pt có 2 nghiệm pb
\(x_1=\dfrac{-3+5}{2.-4};x_2=\dfrac{-3-5}{2.-4}\)
<=> \(x_1=\dfrac{-1}{4}\left(tm\right);x_2=1\left(ktm\right)\)
Vậy nghiệm của pt là : x = -1/4
\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)
\(ĐK:x\ne1\)
\(\Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)
\(\Leftrightarrow-2x^2+x+1=2x^2-2x\)
\(\Leftrightarrow-4x^2+3x+1=0\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow4x^2-4x+x-1=0\)
\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4}\right\}\)
