$\text { Đáp án: }$
$\text { Ta có: }$
` 2a + b = 0 ` ` => ` ` 2a = 0 – b ` ` => ` ` b = 0 – 2a `
` P(–1) . P(3) `
` = [ a.(–1)² + b.(–1) + c ] . [ a.3² + b.3 + c ] `
` = [ a – b + c ] . [ 9a + 3b + c ] `
` = [ a – (0 – 2a) + c ] . [ 9a + 3.(0 – 2a) + c ] `
` = [ a – 0 + 2a + c ] . [ 9a + 0 – 6a + c ] `
` = [ a + 2a + c ] . [ 9a – 6a + c ] `
` = [ 3a + c ] . [ 3a + c ] `
` = (3a + c)² `
$\text { Vì (3a + c)² ≥ 0 }$
` => P(–1) . P(3) ≥ 0 ` $\text { (đpcm) }$