Ta có:

\(\left(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\right)\)

\(=x^2-6x+13-\left(x^2-6x+10\right)\)

\(=3\)

mà  \(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}=1\)

=>   \(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}=3\) 

Em chưa hiểu ở dòng thứ 3,chị có thể giải thích cho em với được ko ạ

a) Đk \(x>0\)và \(x\ne4\)

=\(\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right)\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2}{\sqrt{x}+2}\)

b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)

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

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

Vì \(2\left(\sqrt{x}+2\right)>0\)

mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

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

Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)

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

=\(\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\):\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)=\(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\).\(\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

=\(\frac{-3}{\sqrt{x}+3}\)

câu b c thì sao ạ

\(\sqrt{\sqrt{3}-\sqrt{1-\sqrt{21-12\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(2\sqrt{3}\right)^2-2\cdot2\sqrt{3}\cdot3+3^2}}}\)

\(=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(2\sqrt{3}-3\right)^2}}}\)

\(=\sqrt{\sqrt{3}-\sqrt{1-2\sqrt{3}+3}}\)

\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{\sqrt{3}-\sqrt{3}+1}\)

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