\(a,C=\left(\dfrac{2x+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2+3}{x^2+x+1}\right)\)
\(\Leftrightarrow\left(\dfrac{2x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(\Leftrightarrow\dfrac{2x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\dfrac{x^3-1-x^3-3x+x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\dfrac{-x^2+x}{x^3-1}.\dfrac{x^3-1}{x^2-3x+2}\)
\(\Leftrightarrow\dfrac{-x\left(x-1\right)}{x^2-2x-x+2}\)
\(\Leftrightarrow\dfrac{-x\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\dfrac{-x}{x-2}\)
b, Để C = 3
\(\Leftrightarrow\dfrac{-x}{x-2}=3\)
\(\Leftrightarrow-x=3x-6\)
\(\Leftrightarrow-x-3x=-6\)
\(\Leftrightarrow-4x=-6\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ..................
