1) a)$y=(x+3)(5-x)\leq \frac{(x+3)^2+(5-x)^2}{2}=\frac{2(x^2-1)+32}{2}=16$Vậy $Max=16$ khi $x=1$câu b tương tự
1) a)$y=(x+3)(5-x)\leq \frac{(x+3)^2+(5-x)^2}{2}=\frac{2(x^2-1)+32}{2}=16$Vậy $Max=16$ khi $x=1$
1) a)$y=(x+3)(5-x)\leq \frac{(x+3)^2+(5-x)^2}{2}=\frac{2(x^2-1)+32}{2}=16$Vậy $Max=16$ khi $x=1$
câu b tương tự