Question

[Do Dang Minh]

Giup minh voi : cho P= \(\sqrt{x-1}\)+\(\sqrt{3-x}\)Tim GTLN,GTNN cua P

Answer 1

[Tuyển Trần Thị]

dk \(1\le x\le3\)

\(P^2=x-1+3-x+2\sqrt{\left(x-1\right)\left(3-x\right)}\) =\(2+2\sqrt{\left(x-1\right)\left(3-x\right)}\)

ta co \(p^2\ge2\Rightarrow p\ge\sqrt{2}\) dau = xay ra khi \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

\(P^2=2+2\sqrt{\left(x-1\right)\left(3-x\right)}\le2+x-1+3-x=4\) (ap dung bdt amgm)\(\Rightarrow p\le2\)

dau = xay ra khi \(x-1=3-x\Leftrightarrow x=2\) 

kl min p= \(\sqrt{2}khi\orbr{\begin{cases}x=1\\x=3\end{cases}}\) maxp= 2 khix=2

Answer 2

[cutecuteo]

\(\text{Đ}\text{ể}Pc\text{ó}ngh\text{ĩa}\Leftrightarrow\sqrt{x-1}\ge0\Leftrightarrow x-1\ge0\Leftrightarrow x\ge1\)>=1\(v\text{à}\sqrt{3-x}\ge0\Leftrightarrow3-x\ge0\Leftrightarrow x\le3\).\(x\ge1V\text{à}x\le3\Rightarrow PKh\text{ô}ngC\text{ó}Ngh\text{ĩa}\)