Question

Tìm giá trị lớn nhất      H=(x/x^2-3x+9)-11/x^3+27+1/x+3)÷x^2-1/x+3

Answer 1

`H=(x/(x^2-3x+9)-11/(x^3+27)+1/(x+3)):(x^2-1)/(x+3)`

`H=((x(x+3))/((x+3)(x^2-3x+9))-11/((x+3)(x^2-3x+9))+(x^2-3x+9)/((x+3)(x^2-3x+9))):(x^2-1)/(x+3)`

`H=(x^2+3x-11+x^2-3x+9)/((x+3)(x^2-3x+9))*(x+3)/(x^2-1)`

`H=(2x^2-2)/((x+3)(x^2-3x+9))*(x+3)/(x^2-1)`

`H=2/(x^2-3x+9)`

Vì `x^2-3x+9=x^2-3x+9/4+27/4=(x-3/2)^2+27/4>=27/4>0`

`<=>2/(x^2-3x+9)<=8/27`

Dấu "=" xảy ra khi `x=3/2`

Vậy `max_H=8/27<=>x=3/2`