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

\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{6}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\left|\sqrt{2}-\sqrt{3}\right|\)

Mà\(1< \sqrt{2};\sqrt{2}< \sqrt{3}\)

\(\Rightarrow\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}\)

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

ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)

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

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

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

2) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)

Điều kiện : x>=0

\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)

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

Ta có: A = \(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)

= \(\left|\sqrt{6}-2\sqrt{2}\right|\) \(-\sqrt{18-2.6\sqrt{3}+6}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{\left(\sqrt{18}-\sqrt{6}\right)^2}\)

= \(2\sqrt{2}-\sqrt{6}-\sqrt{18}+\sqrt{6}\)

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

Đặt \(\sqrt[3]{2}=a\)

\(A+\dfrac{a+a^2+a^3}{a^2+a+1}=\dfrac{a\left(a^2+a+1\right)}{a^2+a+1}=a=\sqrt[3]{2}\)

\(P=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}+\sqrt{3+\sqrt{5}}\right)}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}\)

\(=\dfrac{3\sqrt{2}+\sqrt{10}}{4+\sqrt{6+2\sqrt{5}}}+\dfrac{3\sqrt{2}-\sqrt{10}}{4-\sqrt{6-2\sqrt{5}}}\)

\(=\dfrac{3\sqrt{2}+\sqrt{10}}{5+\sqrt{5}}+\dfrac{3\sqrt{2}-\sqrt{10}}{5-\sqrt{5}}\)

\(=\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(5-\sqrt{5}\right)+\left(3\sqrt{2}-\sqrt{10}\right)\left(5+\sqrt{5}\right)}{20}\)

\(=\dfrac{15\sqrt{2}-3\sqrt{10}+5\sqrt{10}-5\sqrt{2}+15\sqrt{2}+3\sqrt{10}-5\sqrt{10}-5\sqrt{2}}{20}\)

\(=\dfrac{30\sqrt{2}-10\sqrt{2}}{20}=\dfrac{20\sqrt{2}}{20}=\sqrt{2}\)

\(\)