a: \(A=\left(\dfrac{3x^2+3}{x^3-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{1}{x-1}\right):\dfrac{2x^2-5x+4}{x-1}\)
\(=\dfrac{3x^2+3-\left(x-1\right)^2-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+4}\)
\(=\dfrac{3x^2+3-x^2-x-1-\left(x^2-2x+1\right)}{x^2+x+1}\cdot\dfrac{1}{2x^2-5x+4}\)
\(=\dfrac{2x^2-x+2-x^2+2x-1}{x^2+x+1}\cdot\dfrac{1}{2x^2-5x+4}=\dfrac{x^2+x+1}{x^2+x+1}\cdot\dfrac{1}{2x^2-5x+4}\)
\(=\dfrac{1}{2x^2-5x+4}\)
b: \(2x^2-5x+4=2\left(x^2-\dfrac{5}{2}x+2\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{7}{16}\right)\)
\(=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{7}{8}>=\dfrac{7}{8}\forall x\) thỏa mãn ĐKXĐ
=>\(A=\dfrac{1}{2x^2-5x+4}< =1:\dfrac{7}{8}=\dfrac{8}{7}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(x-\dfrac{5}{4}=0\)
=>\(x=\dfrac{5}{4}\)
