Question

Cho K = (\(\dfrac{x^2}{x^2-5x+6}\) + \(\dfrac{x^2}{x^2-3x+2}\)) . \(\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
1.Rút gọn K
2.Tìm GTLN của K

Answer 1

1: ĐKXĐ: \(x\notin\left\{1;2;3\right\}\)

\(K=\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)

\(=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-2\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)

\(=\dfrac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^2\left(2x-4\right)}{\left(x-2\right)}\cdot\dfrac{1}{x^4+x^2+1}=\dfrac{2x^2}{x^4+x^2+1}\)