Question

Tìm GTLN của biểu thức sau :

 \(P=3-3x-x^2\)

Admin giúp em nha

Answer 1

P= -(x² + 2.3x/2 + 9/4) +3 +9/4

GTLN: P = 21/4

Answer 2

\(P=-\left(x^2+3x-3\right)=-\left(x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{21}{4}\right)=-\left[\left(x+\frac{3}{2}\right)^2-\frac{21}{4}\right]=-\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\)

Do \(\left(x+\frac{3}{2}\right)^2\ge0,x\in R\)

nên \(-\left(x+\frac{3}{2}\right)^2\le0,x\in R\) 

mà \(-\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\le\frac{21}{4},x\in R\)

VẬy \(Max_P=\frac{21}{4}\)khi \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)