Đặt: \(a=\sqrt{2+x};b=\sqrt{2-x}\left(a,b\ge0\right)\)

\(\Rightarrow\hept{\begin{cases}a^2+b^2=4\\a^2-b^2=2x\end{cases}}\)

\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{4+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{4+ab}\)

\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a-b\right)\left(4+ab\right)}{4+ab}=\sqrt{2+ab}\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=\sqrt{4+2ab}\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=\sqrt{\left(a^2+b^2+2ab\right)}\left(a-b\right)=\left(a+b\right)\left(a-b\right)\)

\(\Rightarrow A\sqrt{2}=a^2-b^2=2x\)

\(\Rightarrow A=x\sqrt{2}\)

1.

$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$

$=x+3+(3-x)=6$

2.

$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

4.

$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$

$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$

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

\(=\dfrac{x-\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)

b: Để A<=3/căn x thì \(\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}< =\dfrac{3}{\sqrt{x}}\)

=>\(\dfrac{x-2\sqrt{x}-1-3x+6\sqrt{x}-3}{\left(\sqrt{x}-1\right)^2}< =0\)

=>\(-2x+4\sqrt{x}-4< =0\)

=>\(x-2\sqrt{x}+2>=0\)(luôn đúng)

Đặt \(a=\sqrt{2+x};\text{ }b=\sqrt{2-x}\Rightarrow a^2+b^2=4\)

\(A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}=\left(a-b\right)\sqrt{\frac{a^2+b^2}{2}+ab}\)

\(=\left(a-b\right)\sqrt{\frac{\left(a+b\right)^2}{2}}=\frac{\left(a-b\right)\left(a+b\right)}{\sqrt{2}}\)

\(=\frac{a^2-b^2}{\sqrt{2}}=\frac{\left(2+x\right)-\left(2-x\right)}{\sqrt{2}}=\frac{2x}{\sqrt{2}}=x\sqrt{2}\)

\(\hept{\begin{cases}-1\le x\le1\\2-\sqrt{1-x^2}\end{cases}\Rightarrow-1\le x\le1\left(^∗\right)}\)

Đặt : \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}\Rightarrow\hept{\begin{cases}a^2+b^2=2\\a,b\ge0\end{cases}}}\)

A tồn tại mọi x thuộc ( * )

\(A=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{a^2-2ab+b^2}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)

\(A=\frac{\sqrt{2}\sqrt{\left(a-b\right)^2}\left(a+b\right)\left(2-ab\right)}{\left(2-ab\right)}\) . Vói đk ( \(I\)) \(A=\frac{\sqrt{2}}{2}!a-b!\left(a+b\right)\)

\(\orbr{\begin{cases}\hept{\begin{cases}a\ge b\Leftrightarrow0\le x\le1\\A=\frac{\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{\sqrt{2}}{2}x\end{cases}}\\\hept{\begin{cases}a< b\Leftrightarrow-1\le x< 0\\A=\frac{-\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{-\sqrt{2}}{2}x\end{cases}}\end{cases}}\)

\(\Rightarrow A=\frac{\sqrt{2}}{2}!x!\) . Với x thỏa mãn điều kiện ( * )

\(y=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

\(\Rightarrow y^2=2x+2\sqrt{x+2\sqrt{x-1}}.\sqrt{x-2\sqrt{x-1}}\)

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

\(\Rightarrow y=2\)