Đáp án:
`a)`
` x^2 + x +1 = (x^2 + 1/(2).2.x + 1/4) - 1/4 + 1`
` = (x -1/2)^2 + 3/4`
Ta có
` (x-1/2)^2 \ge 0`
` => (x-1/2)^2 + 3/4 \ge 3/4`
` => (x-1/2)^2 +3/4 > 0`
` => x^2 + x + 1 > 0`
`b)`
` -4x^2 - 4x - 2 = - (4x^2 + 4x +1) - 1`
` = -(2x +1)^2 - 1`
Ta có ` -(2x+1)^2 \le 0`
` => - (2x+1)^2 - 1 \le -1 < 0`
` => -4x^2 - 4x - 2 < 0`
`c)`
` x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15`
` = (x^2 - 2x +1) + 4.(y^2 + 2y +1) + (z^2 - 6z + 9) + 1`
` = (x-1)^2 + 4(y+1)^3 + (z -3)^2 + 1`
Ta có
` (x-1)^2 \ge 0 ; (y+1)^2 \ge 0 ; (z-3)^2 \ge 0`
` => (x-1)^2 + 4(y+1)^3 + (z -3)^2 + 1 \ge 1 > 0`
` => x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15 > 0`
