Đáp án:
`A=6/(6x-5-9x^2)=6/(-(9x^2-6x+5))`
`=6/(-(9x^2-6x+1+4))=6/(-(3x-1)^2+4)`
Ta có : `(3x-1)^2>=0 \ \ ∀x`
`to -(3x-1)^2<= 0`
`to -(3x-1)^2-4 <= -4`
`to 6/(-(3x-1)^2+4) >= 6/-4 = -3/2`
Dấu "=" xảy ra khi : `(3x-1)^2=0`
`<=> x=1/3`
Vậy `A_(min)=-3/2 <=> x=1/3`