\(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\sqrt{\left(x-1\right)^2}+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|x-6\right|\right)\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\)

Ta có \(\hept{\begin{cases}\left|x-4\right|\ge0\forall x\\\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\end{cases}}\)

=> \(\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\ge5\forall x\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-4=0\\\left(x-1\right)\left(6-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

=> MinA = 5 <=> x = 4

Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(\Rightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left|x-1\right|+\left|x-6\right|\)

Xét \(\left|x-1\right|+\left|x-6\right|\)ta có: 

\(\left|x-1\right|+\left|x-6\right|=\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\)(1)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(6-x\right)\ge0\)

TH1: Nếu \(\hept{\begin{cases}x-1< 0\\6-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\6< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>6\end{cases}}\)( vô lý )

TH2: Nếu \(\hept{\begin{cases}x-1\ge0\\6-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\6\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le6\end{cases}}\Leftrightarrow1\le x\le6\)

mà \(\left|x-4\right|\ge0\)(2)

Từ (1) và (2) \(\Rightarrow A\ge5\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-4=0\\1\le x\le6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

Vậy \(minA=5\)\(\Leftrightarrow x=4\)

Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

    \(\Leftrightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

    \(\Leftrightarrow A=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

Vì \(\left|a\right|=\left|-a\right|\) \(\Rightarrow\)\(\left|x-6\right|=\left|6-x\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:

     \(\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=5\)

      \(\Rightarrow\)\(A\ge\left|x-4\right|+5\)

Vì \(\left|x-4\right|\ge0\forall x\)\(\Rightarrow\)\(\left|x-4\right|+5\ge5\forall x\)

      \(\Rightarrow\)\(A\ge5\)

Dấu "=" xảy ra khi:  \(\hept{\begin{cases}\left(x-1\right)\left(6-x\right)>0\\x-4=0\end{cases}}\)

                         \(\Leftrightarrow\hept{\begin{cases}1< x< 6\\x=4\end{cases}}\)

                           \(\Rightarrow x=4\)

Vậy \(A_{min}=5\)\(\Leftrightarrow\)\(x=4\)

Chắc dùng Mincowski

ĐKXĐ: x>=0

a: P=1/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{1}{2}\)

=>\(2\sqrt{x}+4=\sqrt{x}+5\)

=>\(\sqrt{x}=1\)

=>x=1(nhận)

b: \(P^2-P=P\left(P-1\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\cdot\dfrac{\sqrt{x}+2-\sqrt{x}-5}{\sqrt{x}+5}\)

\(=\dfrac{-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+5\right)^2}< 0\)

=>\(P^2< P\)

c: Để P nguyên thì \(\sqrt{x}+2⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5-3⋮\sqrt{x}+5\)

=>\(\sqrt{x}+5\inƯ\left(-3\right)\)

=>\(\sqrt{x}+5\in\left\{1;-1;3;-3\right\}\)

=>\(\sqrt{x}\in\left\{-4;-6;-2;-8\right\}\)

=>\(x\in\varnothing\)