Question

Tìm Max A=\(\sqrt{x-1}+\sqrt{9-x}\)

Answer 1

Nhận xét : A > 0

Áp dụng bđt Bunhiacopxki , ta có : 

\(A^2=\left(1.\sqrt{x-1}+1.\sqrt{9-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+9-x\right)\)

\(\Rightarrow A^2\le16\Rightarrow A\le4\)

Suy ra Max A = 4 <=> \(\begin{cases}1\le x\le9\\\sqrt{x-1}=\sqrt{9-x}\end{cases}\) \(\Leftrightarrow x=5\)

 

Answer 2

\(A^2=\left(x-1\right)+\left(9-x\right)+2\sqrt{\left(x-1\right)\left(9-x\right)}\)

\(=8+2\sqrt{\left(x-1\right)\left(9-x\right)}\).Dùng BĐT cô-si

\(=8+2\sqrt{\left(x-1\right)\left(9-x\right)}\le8+\left(x-1\right)+\left(9-x\right)=16\)

\(\Rightarrow A^2\le16\Leftrightarrow A\le4\)

Dấu = khi \(\begin{cases}1\le x\le9\\\sqrt{x-1}=\sqrt{9-x}\end{cases}\)\(\Leftrightarrow x=5\)

Vậy MaxA=4 khi x=5