\(CM:\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)=2x+1\)
(ĐKXĐ: x ≠ 0; x ≠ 1 )
*Biến đổi vế trái :
\(\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)\)
\(=\dfrac{x^3-1}{x}\left(\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=\dfrac{x^3-1}{x}.\dfrac{\left(x+1\right)\left(x-1\right)+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-1}{x}.\dfrac{x^2-1+x^2+x+1}{x^3-1}\)
\(=\dfrac{2x^2+x}{x}=\dfrac{x\left(2x+1\right)}{x}\)
\(=2x+1\)
Vậy:
\(\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)=2x+1\)
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