ĐKXĐ: \(x\ne1\)
\(\dfrac{1}{x-1}-\dfrac{3}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+2x+1\right)}\)
\(\Rightarrow x^2+x+1-3=2x\left(x-1\right)\)
\(\Leftrightarrow x^2+x-2=2x^2-2x\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
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