Tìm GTLN: -4x^2- 12x

Đăth $A=-4x^2-12x$

$A=-4.(x^2+3x)$

$=-4.(x^2+2.x..\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4})$

$=-4.(x+\dfrac{3}{2})^2+9$

Ta thấy: $(x+\dfrac{3}{2})^2≥0$

$→-4(x+\dfrac{3}{2})^2≤0$

$→-4(x+\dfrac{3}{2})^2+9≤9$

$→$ Dấu "=" xảy ra khi $x+\dfrac{3}{2}=0$

$→x=-\dfrac{3}{2}$

$→A_{max}=9$

Vậy $A_{max}=9$ khi $x=-\dfrac{3}{2}$