Đáp án+Giải thích các bước giải:
`1) 9x^2 + 16 ≥ 16`
Vì ` 9x^2 ≥ 0 ∀ x in RR`
nên `9x^2 + 16 ≥ 16 ∀ x in RR`
Vậy `9x^2 + 16 ≥ 16`
`2) (36-x)^2-25 ≥ -25`
Vì `(36-x)^2 ≥ 0 ∀ x in RR`
nên `(36-x)^2-25 ≥ -25 ∀ x in RR`
Vậy `(36-x)^2-25 ≥ -25`
`3) 4x^2 - 4x+1 ≥ 0`
Ta có: `4x^2 - 4x +1`
`= (2x)^2 - 2.2x.1+1^2`
`= (2x-1)^2 ≥ 0 `
Vậy `4x^2 - 4x+1 ≥ 0`
`4) 9x^2 - 12x + 5 >0`
Ta có: `9x^2 - 12x +5`
`= (3x)^2 - 2.3x.2 + 2^2 +1`
`= (3x - 2)^2 +1`
Vì `(3x - 2)^2≥ 0 ∀ x in RR`
nên `(3x - 2)^2 +1 ≥ 1 > 0 ∀ x in RR`
Vậy `9x^2 - 12x + 5 >0`
`5) 4x^2 + 12x +10 >0`
Ta có: `4x^2 + 12x +10`
`= (2x)^2 + 2.2x.3 + 3^2 +1`
`= (2x+3)^2 +1`
Vì `(2x+3)^2≥ 0 ∀ x in RR`
nên `(2x+3)^2 +1 ≥ 1 > 0 ∀ x in RR`
Vậy `4x^2 + 12x +10 >0`
`6) x^2 -x +1 >0`
Ta có: `x^2 -x +1`
`= x^2 + 2.x . 1/2 + 1/4 + 3/4`
`= (x+1/2)^2 + 3/4`
Vì `(x+1/2)^2 ≥ 0 ∀ x in RR`
nên `(x+1/2)^2 + 3/4 ≥ 3/4 > 0 ∀ x in RR`
Vậy `x^2 -x +1 >0`
`7) 4x - 4x^2 - 2 < 0`
Ta có: `4x - 4x^2 - 2`
`= -(4x^2 - 4x + 2)`
`= -[(2x)^2 - 2.2x.1 +1^2 +1]`
`= -(2x - 1)^2 -1`
Vì `-(2x - 1)^2 ≤ 0 ∀ x in RR`
nên `-(2x - 1)^2 -1 ≤ -1 <0 ∀ x in RR`
Vậy `4x - 4x^2 - 2 < 0`
`8) 10x - 25x^2 - 3 < 0`
Ta có: `10x - 25x^2 - 3`
`= -(25x^2 - 10x + 3)`
`=-[(5x)^2 - 2.5x.1 +1^2 +2]`
`= -(5x - 1)^2 - 2`
Vì `-(5x - 1)^2 ≤ 0∀ x in RR`
nên `-(5x - 1)^2 - 2 ≤ -2 < 0∀ x in RR`
Vậy `10x - 25x^2 - 3 < 0`
