a, ĐK: \(x\ge4;x\le-4\)

\(\sqrt{x^2-4-12}\le x-4\)

\(\Leftrightarrow\sqrt{x^2-16}\le x-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\x^2-16\le\left(x-4\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-16\le x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow x=4\left(tm\right)\)

b, ĐK: \(x\ge8;x\le0\)

\(\sqrt{x^2-8x}\ge2\left(x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\left(x+1\right)\ge0\\x^2-8x\ge4\left(x^2+2x+1\right)\end{matrix}\right.\\2\left(x+1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow x\le\dfrac{-8+2\sqrt{13}}{3}\)

c, \(\left(x-2\right)\sqrt{x^2+4}< x^2-4\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-\sqrt{x^2+4}\right)>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\x+2-\sqrt{x^2+4}< 0\end{matrix}\right.\left(I\right)\text{v}\left\{{}\begin{matrix}x-2< 0\\x+2-\sqrt{x^2+4}>0\end{matrix}\right.\left(II\right)\)

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}x>2\\x+2< \sqrt{x^2+4}\end{matrix}\right.\Leftrightarrow...\)

\(\left(II\right)\Leftrightarrow\left\{{}\begin{matrix}x-2< 0\\x+2-\sqrt{x^2+4}>0\end{matrix}\right.\Leftrightarrow...\)

1) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

= \(\frac{ \left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

= \(\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\) = \(\frac{\left(\sqrt{7}\right)^2+2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}{\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2}\)

= \(\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\) = \(\frac{24}{2}=12\)

2) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)

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

= \(x+2y-\left(x-2y\right)\) = \(x+2y-x+2y=4y\)

3)\(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)

= \(4x+x-12=5x-12\)

a/ \(\sqrt{4a^2}=\sqrt{\left(2a\right)^2}=\left|2a\right|=2a\)

b/ \(\sqrt{\left(\frac{2}{5}\right)^2\left(x-2\right)^2}=\frac{2}{5}\left|x-2\right|=\frac{2}{5}\left(x-2\right)=\frac{2x}{5}-\frac{4}{5}\)

c/ \(\sqrt{5^2\left(3-a\right)^2}+3=5\left|3-a\right|+3=\left[{}\begin{matrix}18-5a\left(a\le3\right)\\5a-12\left(a\ge3\right)\end{matrix}\right.\)

d/ \(=\frac{1}{2\left(x-5\right)}.6\left|x-5\right|=\frac{3\left|x-5\right|}{x-5}=\left[{}\begin{matrix}3\left(x>5\right)\\-3\left(x< 5\right)\end{matrix}\right.\)

a) \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

b) \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}+1-\sqrt{5}+1\)

\(=2\)

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

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

\(=-\left|x-2\right|\)

\(=-x+2\)

\(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|=\sqrt{5}+1-\sqrt{5}+1=2\)

a: 

ĐKXĐ: x>=5/2

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)

=>\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\cdot\sqrt{2x-5}}=14\)

=>\(\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

=>\(\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)

=>\(2\sqrt{2x-5}+4=14\)

=>\(\sqrt{2x-5}=5\)

=>2x-5=25

=>2x=30

=>x=15

b: \(x^2-4x=\sqrt{x+2}\)

=>\(x+2=\left(x^2-4x\right)^2\) và x^2-4x>=0

=>x^4-8x^3+16x^2-x-2=0 và x^2-4x>=0

=>(x^2-5x+2)(x^2-3x-1)=0 và x^2-4x>=0

=>\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\)

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

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

\(=\left(\sqrt{x-2}+\sqrt{2}\right)^2=VP\left(\text{đ}pcm\right)\)