Sửa đề: \(\dfrac{x^2-2x+2}{x^2-x+1}-\dfrac{x^2}{x^2+x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
ĐKXĐ: \(x\notin\left\{0\right\}\)
Ta có: \(\dfrac{x^2-2x+2}{x^2-x+1}-\dfrac{x^2}{x^2+x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\dfrac{\left(x^2-2x+2\right)\left(x^2+x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}-\dfrac{x^2\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\dfrac{x^4+x^3+x^2-2x^3-2x^2-2x+2x^2+2x+2-x^4+x^3-x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\dfrac{2x}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
Suy ra: 2x=3
hay \(x=\dfrac{2}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{2}{3}\right\}\)
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