\(A^2=\left(\sqrt{1-x}+\sqrt{1+x}\right)^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)=4\\ \Leftrightarrow A\le2\\ A_{max}=2\Leftrightarrow1-x=1+x\Leftrightarrow x=0\\ A^2=2+2\sqrt{1-x^2}\ge2\\ \Leftrightarrow A\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow1-x^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(\sqrt{2}\le A\le2\)

\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)

\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)

\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)

\(A_{min}=6\) khi \(x=5\)

1:

a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

Lời giải:
TXĐ: $[-1;1]$

$y'=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{1-x}}+\frac{x}{2}$

$y'=0\Leftrightarrow x=0$

$f(0)=2$;

$f(1)=f(-1)=\sqrt{2}+\frac{1}{4}$
Vậy $f_{\min}=2; f_{\max}=\frac{1}{4}+\sqrt{2}$

\(B=\dfrac{x-\sqrt[]{x}}{\sqrt[]{x}-\left(x+1\right)}\)

\(B\) xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-\left(x+1\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x+1\ne0,\forall x\in R\end{matrix}\right.\) \(\Leftrightarrow x\ge0\)

\(\Leftrightarrow B=\dfrac{x-\sqrt[]{x}+1-1}{-\left(x-\sqrt[]{x}+1\right)}\)

\(\Leftrightarrow B=-1+\dfrac{1}{x-\sqrt[]{x}+1}\)

\(\Leftrightarrow B=-1+\dfrac{1}{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+1}\)

\(\Leftrightarrow B=-1+\dfrac{1}{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

mà \(\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\ge0\)

\(\Rightarrow B=-1+\dfrac{1}{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le-1+\dfrac{4}{3}=\dfrac{1}{3}\)

\(\Rightarrow GTLN\left(B\right)=\dfrac{1}{3}\left(tại.x=\dfrac{1}{4}\right)\)