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

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

x=1 hoặc x= -1

mình ra là x=1 hoawcjx=-1

https://vn.answers.yahoo.com/question/index?qid=20111127061818AAhrlFU

vào đây xem nhé

pt\(\Leftrightarrow\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-4}}{y}\)

Áp dụng BĐT cô si cho 2 số ko âm ta có:

\(\sqrt{x-1}=\sqrt{1\left(x-1\right)}\le\frac{x+1-1}{2}=\frac{x}{2}\)

\(\Rightarrow\frac{\sqrt{x-1}}{x}\le\frac{1}{2}\)(vì x dương)

\(\sqrt{y-4}=\frac{1}{2}\sqrt{4\left(y-4\right)}\le\frac{1}{2}.\frac{4+y-4}{2}=\frac{y}{4}\)

\(\Rightarrow\frac{\sqrt{y-4}}{y}\le\frac{1}{4}\)(vì y dương)

\(\Rightarrow Q=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-4}}{y}\le\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

Vậy \(Q\)max là \(\frac{3}{4}\)khi \(x=2,y=8\)

nx \(x^2-x+4=2\sqrt{x^2+3}\le\frac{2^2+x^2+3}{2}=\frac{x^2+7}{2}\) (am gm)

\(\Rightarrow x^2-x+4\le\frac{x^2+7}{2}\Leftrightarrow2x^2-2x+8\le x^2+7\Leftrightarrow x^2-2x+1\le0\)

                                                         \(\Leftrightarrow\left(x-1\right)^2\le0\) ma \(\left(x-1\right)^2\ge0\) 

                      \(\Rightarrow x=1\)

\(Pt\Leftrightarrow\left(\sqrt{x^2+3}-1+\sqrt{x}\right)\left(\sqrt{x^2+3}-1-\sqrt{x}\right)=0\)