\(\dfrac{3x^2+5x+1}{x^3-1}-\dfrac{1-x}{x^2+x+1}-\dfrac{3}{x+1}\)
\(=\dfrac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2+3x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
=\(\dfrac{3x^2+5x+1+x^2-2x+1-3x^2-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+5}{x^3-1}\)
\(\dfrac{3x^2+5x+1}{x^3-1}-\dfrac{1-x}{x^2+x+1}-\dfrac{3}{x-1}\)
=\(\dfrac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
=\(\dfrac{\left(3x^2+5x+1\right)-\left(x-1\right)\left(1-x\right)-3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
=\(\dfrac{3x^2+5x+1-x+1-3x^2-3x-3}{\left(x^2+x+1\right)}\)
=\(\dfrac{x-1}{x^2+x+1}\)
Đúng thì tick giúp mình nhé!
