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a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)
\(A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(B=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\)
\(C=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(7-5\right)=2\sqrt{9}\)
\(D=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(E=\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{5^2-\sqrt{5}^2}=\dfrac{60}{20}=3\)
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
1: \(=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
2: \(=\dfrac{4+2\sqrt{3}+4-2\sqrt{3}}{2}=\dfrac{8}{2}=4\)
4: \(=\dfrac{-3+5\sqrt{3}}{11}+\dfrac{3+5\sqrt{3}}{11}=\dfrac{10\sqrt{3}}{11}\)
a: \(=\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
d: \(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)
\(1a.\left(\sqrt{72}-3\sqrt{5}+2\sqrt{8}\right).\sqrt{2}+\sqrt{90}=\sqrt{144}-3\sqrt{10}+2.\sqrt{16}+3\sqrt{10}=12+8=20\) \(b.\left(\sqrt{\dfrac{1}{5}}-10\sqrt{\dfrac{27}{5}}+2\sqrt{5}\right):\sqrt{5}+6\sqrt{3}=\left(\sqrt{\dfrac{1}{5}}-30\sqrt{\dfrac{3}{5}}+2\sqrt{5}\right).\dfrac{1}{\sqrt{5}}+6\sqrt{3}=\dfrac{1}{5}-6\sqrt{3}+2+6\sqrt{3}=\dfrac{11}{5}\) \(2.\sqrt{\left(3-\sqrt{10}\right)^2}=\sqrt{10}-3\)
\(b.\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}=2+\sqrt{3}+2-\sqrt{3}=4\) \(c.\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)
1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha 
@Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé 
a: \(=\left(\dfrac{\sqrt{2}}{4}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\cdot10\sqrt{2}\right)\cdot8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}\)
\(=54\sqrt{2}\)
b: \(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)
c: \(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
d: \(=\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}=0\)
\(a.\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\dfrac{\left(2+\sqrt{3}\right)\sqrt{3-2\sqrt{3}+1}}{\sqrt{3+2\sqrt{3}+1}}=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{2\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3-1}=4-3=1\)
\(b.\dfrac{\left(\sqrt{5}-1\right)^3}{\sqrt{5}-2}=\dfrac{5\sqrt{5}-15+3\sqrt{5}-1}{\sqrt{5}-2}=\dfrac{8\sqrt{5}-16}{\sqrt{5}-2}=\dfrac{8\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=8\)
\(c.\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right]=2\left(3+1+3\right)=2.7=14\)
\(d.\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=\dfrac{2}{2}=1\)
Q=\(\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{2}+1\right)+\sqrt{3}\left(2+\sqrt{2}\right)}{\sqrt{3}\left(\sqrt{2}+1\right)}-\left(\sqrt{2}+\sqrt{3}\right)\)
Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)-\(\left(\sqrt{2}+\sqrt{3}\right)\)
Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}-2\sqrt{3}-3\sqrt{2}-\sqrt{6}-3}{\sqrt{3}\left(\sqrt{2}+1\right)}\)
Q=\(\dfrac{2\sqrt{3}-2\sqrt{6}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)=\(\dfrac{\sqrt{4}-\sqrt{8}}{\sqrt{2}+1}\)