A = \(\sqrt{3+2\sqrt{2}}-\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)

   = \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{3+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}+3}\)

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

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

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

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

   = \(-\sqrt{3}\)

\(a,\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{2}\right)}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{6}}\left(\dfrac{2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-3\sqrt{2}}{4-6}\right)-\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}-1}{2+\sqrt{6}}-\dfrac{\sqrt{2}-\sqrt{3}}{2\sqrt{2}.\sqrt{3}}.\dfrac{4\sqrt{3}}{-2}-\dfrac{1}{\sqrt{2}}\)

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

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

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

\(=\dfrac{\sqrt{2}+\sqrt{3}-1+2+\sqrt{6}-\sqrt{6}-3-\sqrt{2}-\sqrt{3}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}\)

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

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

.

Lời giải:

$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$

\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)

Do đó:

\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)

\(A=\frac{\text{TS}}{\text{MS}}=1\)

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

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

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

\(=2\sqrt{2}-\frac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}\)

\(=2\sqrt{2}-\frac{6\sqrt{2}}{6}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)

\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)

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

\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)

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

\(=4\)

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

\(=\dfrac{-2\sqrt{2}+\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}+\sqrt{6+3\sqrt{3}}}{\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{2}\left(\sqrt{3}-1\right)-\sqrt{6}-\sqrt{6-3\sqrt{3}}}{\sqrt{3}}\)

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

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

b) \(\sqrt{\dfrac{1-\sqrt{3}}{1+\sqrt{3}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}}\)

\(=\sqrt{\dfrac{1-\sqrt{3}}{1+\sqrt{3}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}}\)

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

\(=\sqrt{\dfrac{4\sqrt{3}}{2}}=\sqrt{2\sqrt{3}}\)

chịu,,, chắc toàn dấu cộng chứ tự nhiên có dấu trừ sao làm

nếu là dấu cộng cx khó đấy

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)