\(\dfrac{1}{\sqrt{49+20\sqrt{6}}}-\dfrac{1}{\sqrt{49-20\sqrt{6}}}+\dfrac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{5^2+2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{5^2-2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)

\(=\dfrac{1}{\sqrt{\left(5+2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{\left(5-2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{\left(2-\sqrt{3}\right)^2}}\)

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

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

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

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

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

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

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

\(=\dfrac{1}{\sqrt{\sqrt{25}^2+2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}-\dfrac{1}{\sqrt{\sqrt{25}^2-2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}+\dfrac{1}{\sqrt{\sqrt{4}^2-2\sqrt{4}\sqrt{3}+\sqrt{3}^2}}\\ =\dfrac{1}{\sqrt{\left(\sqrt{25}+\sqrt{24}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{25}-\sqrt{24}\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{4}-\sqrt{3}\right)^2}}\\ =\dfrac{1}{5+\sqrt{24}}-\dfrac{1}{5-\sqrt{24}}+\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{5-\sqrt{24}}{25-24}-\dfrac{5+\sqrt{24}}{25-24}+\dfrac{2+\sqrt{3}}{4-3}\\ =5-\sqrt{24}-5-\sqrt{24}+2+\sqrt{3}\\ =2-4\sqrt{6}+\sqrt{3}\)

câu đầu có \(3-12\sqrt{6}< 0\) nên không căn được nên đề bạn sai

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{4^2-2.4.\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.3+3^2}\)

\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}=\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

\(=4-\sqrt{15}+\sqrt{15}-3=1\)

\(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}=\sqrt{49-20\sqrt{6}}-\sqrt{49+20\sqrt{6}}\)

\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}-\sqrt{5^2+2.5.4\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

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

\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=4-\sqrt{15}+\sqrt{15}-3\)

=1

\(\frac{49^6\times5-7^{11}}{\left(-7\right)^{10}\times5+2\times49^5}\)

\(=\frac{7^{12}\times5-7^{11}}{7^{10}\times5+2\times7^{10}}\)

\(=\frac{7^{11}.\left(5-1\right).7}{7^{10}.\left(5+2\right)}\)

\(=\frac{7^2.4}{6}\)

\(=\frac{55}{6}\)

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

\(49+20\sqrt{6}=25+2.5.2\sqrt{6}+24=\left(5+2\sqrt{6}\right)^2=\left(3+2.\sqrt{3}\sqrt{2}+2\right)^2=\left(\sqrt{3}+\sqrt{2}\right)^4\)

\(\Leftrightarrow\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

tuiwng tự \(\Leftrightarrow\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\)

=> Cộng lại  = > dpcm

Ta có \(\sqrt[4]{49+20\sqrt{6}}=\sqrt[4]{25+10\sqrt{24}+24}=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}\)

                               \(=\sqrt[4]{\left(\sqrt{3}+\sqrt{2}\right)^4}=\sqrt{3}+\sqrt{2}\)

Tương tự : \(\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\) ( Do \(\sqrt{3}>\sqrt{2}\) )

Suy ra \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

54 : 6 + 73 = 9+ 73

                  = 82  (0,5điểm)

49 + 6 x 6 = 49 + 36

                 = 85  (0,5 điểm)

a: =-2/49-5/3+2/49=-5/3

b: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)