ĐK : x khác 1
pt <=> \(\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
=> 3x2 + x - 4 = 4x - 4
<=> 3x2 - 3x = 0
<=> 3x( x - 1 ) = 0
<=> x = 0 (tm) hoặc x = 1(loại)
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