Question

(x-1)(x-3)(x-4)(x-6)+10≥1

Answer 1

`(x^2 - 7x + 6)(x^2 - 7x + 12) + 10`

Đặt `x^2 - 7x + 9 = t`.

`-> (t-3)(t+3) + 10 = t^2 - 9 + 10 = t^2 + 1 >= 1`.

Dấu bằng xảy ra `<=> x^2 - 7x + 9 = 0 <=> x = (7+-sqrt 13)/2`

Answer 2

`(x-1)(x-3)(x-4)(x-6)+10 >= 1`

`<=>[(x-1)(x-6)][(x-3)(x-4)] >= -9`

`<=>(x^2-7x+6)(x^2-7x+12) >= -9`

`<=>(x^2-7x+6)^2+6(x^2-7x+6)+9 >= 0`

`<=>(x^2-7x+6+3)^2 >= 0`

`<=>(x^2-7x+9)^2 >= 0` (LĐ `AA x`)