`a) (x - 1)(x + 2)(x + 3)(x + 6)`
`= [(x - 1)(x + 6)].[(x + 2)(x + 3)]`
`= (x² + 5x - 6)(x² + 5x + 6)`
`= (x² + 5x)² - 36`
Ta có:
`(x² + 5x)² - 36 ≥ -36` với `∀ x ∈ RR`
Dấu "=" xảy ra:
`<=> x(x + 5) = 0`
`<=>` \(\left[ \begin{array}{l}x = 0\\x = -5\end{array} \right.\)
`b) x² - 2x + y² + 4y + 8`
`= x² - 2x + 1 + y² + 4y + 4 + 3`
`= (x - 1)² + (y + 2)² + 3`
Ta có;
`(x - 1)² + (y + 2)² + 3 ≥ 3` với `∀ x ∈ RR`
Dấu "=" xảy ra:
`<=>` \(\left\{ \begin{array}{l}x - 1 = 0\\y + 2 = 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x = 1\\y = -2\end{array} \right.\)
`c) x² - 4x + y² - 8y + 6`
`= x² - 4x + 4 + y² - 8y + 16 - 14`
`= (x - 2)² + (y - 4)² - 14`
Ta có:
`(x - 2)² + (y - 4)² - 14 ≥ -14` với `∀ x ∈ RR`
Dấu "=" xảy ra:
`<=>` \(\left\{ \begin{array}{l}x - 2 = 0\\y - 4 =0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x = 2\\y = 4\end{array} \right.\)
