Question

Answer 1

đkxđ : x khác 1 

\(x^2-5x+6\ge0\)

<=> \(x^2-2x-3x+6\ge0\)

<=> \(x\left(x-2\right)-3\left(x-2\right)\ge0\)

<=> \(\left(x-2\right)\left(x-3\right)\ge0\)

=> \(x\ge2;x\ge3\)