Question

tìm giá trị lớn nhất \(\sqrt{x-3}+\sqrt{y-4}\)biết x+y=8

Answer 1

Gọi \(A=\sqrt{x-3}+\sqrt{y-4}\)

Ta có : \(A^2=x-3+y-4=2\sqrt{\left(x-3\right)\left(y-4\right)}=x+y-7+2\sqrt{2\left(x-3\right)\left(y-4\right)}\)

\(=1+2\sqrt{\left(x-3\right)\left(y-4\right)}\)

Theo AM - GM ta có : \(2\sqrt{\left(x-3\right)\left(y-4\right)}\le x-3+y-4=x+y-7=8-7=1\)

\(\Rightarrow A^2\le1+1=2\Rightarrow A\le\sqrt{2}\)Có GTLN là \(\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=y-4\Leftrightarrow\hept{\begin{cases}x-y=-1\\x+y=8\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{2}\\y=\frac{9}{2}\end{cases}}}\)