Bài `3:`
`a, x^2 - 4x + 5 > 0`
Ta có: `x^2 - 4x + 5`
`= x^2 - 4x + 4 + 1`
`= x^2 - 2. x. 2 + 2^2 + 1`
`= (x - 2)^2 + 1`
Ta thấy `(x - 2)^2 ≥ 0 ∀ x` nên `⇒ (x - 2)^2 + 1 > 0 ∀ x` (đpcm)
`b, x^2 + x + 1 > 0`
Ta có: `x^2 + x + 1`
`= x^2 + x + 1/4 + 3/4
`= x^2 + 2. x. 1/2 + (1/2)^2 + 3/4`
`= (x + 1/2)^2 + 3/4`
Ta thấy `(x + 1/2)^2 ≥ 0 ∀ x` nên `⇒ (x + 1/2)^2 + 3/4 > 0 ∀ x` (đpcm)
Bài `5:`
`VT = x^2 + (x - a)^2 + (x - b)^2 + (x - c)^2`
`= x^2 + x^2 - 2xa + a^2 + x^2 - 2xb + b^2 + x^2 - 2xc + c^2`
`= 4x^2 + a^2 + b^2 + c^2 - 2xa - 2xb - 2xc`
`= 2x. 2x + a^2 + b^2 + c^2 - 2x(a + b + c)`
`= 2x(a + b + c) - 2x(a + b + c) + a^2 + b^2 + c^2`
`= a^2 + b^2 + c^2 = VP` (đpcm)