Câu 4:
c) ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x^3-1\right)\left(x^3+1\right)}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2.\left(x^2-x+1\right).\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}-\dfrac{\left(x-1\right)^2.\left(x^2+x+1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Rightarrow\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x+2\right)^2\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^3+1-x^3+1\right)=2\left(x+2\right)^2\)
\(\Leftrightarrow2\left(x^2-1\right)=2\left(x+2\right)^2\)
\(\Leftrightarrow x^2-1=x^2+4x+4\)
\(\Leftrightarrow4x+5=0\)
\(\Leftrightarrow x=\dfrac{-5}{4}\left(nhận\right)\)
-Vậy \(S=\left\{\dfrac{-5}{4}\right\}\)
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