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a, \(3\sqrt{3}\) >\(2\sqrt{3}\) =>\(3\sqrt{3}\) >\(\sqrt{12}\)
b,có \(3\sqrt{5}=\sqrt{45}\) <\(\sqrt{49}=7\) =>7 >\(3\sqrt{5}\)
c,\(\sqrt{\dfrac{51}{9}}\) <\(\sqrt{6}\) => \(\dfrac{1}{3}\sqrt{51}\) <\(\dfrac{1}{5}\sqrt{150}\)
d.\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
a/ \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=7-4\sqrt{3}+7+4\sqrt{3}=14\)
a) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}+2}\right)-\dfrac{12}{3-\sqrt{6}}\)
\(=\left(\dfrac{15\left(\sqrt{6}+2\right)+4\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}+2\right)}\right)-\dfrac{12}{3-\sqrt{6}}=\dfrac{15\sqrt{6}+30+4\sqrt{6}+4}{6+2\sqrt{6}+\sqrt{6}+2}-\dfrac{12}{3-\sqrt{6}}\) \(=\dfrac{34+19\sqrt{6}}{8+3\sqrt{6}}-\dfrac{12}{3-\sqrt{6}}=\dfrac{\left(34+19\sqrt{6}\right)\left(3-\sqrt{6}\right)-12\left(8+3\sqrt{6}\right)}{\left(8+3\sqrt{6}\right)\left(3-\sqrt{6}\right)}\)
\(=\dfrac{102-34\sqrt{6}+57\sqrt{6}-114-96-36\sqrt{6}}{24-8\sqrt{6}+9\sqrt{6}-18}=\dfrac{-108-13\sqrt{6}}{6+\sqrt{6}}\)
c) \(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{3}}=2\sqrt{2+\sqrt{3}}=\sqrt{2}.\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
câu này mk cảm thấy đề sai thì phải ; mà nếu o phải đề sai thì lời giải đó nha
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
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. Ta có \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
Vậy \(3\sqrt{3}>\sqrt{12}\)
b. Ta có \(7=\sqrt{49}\), \(3\sqrt{5}=\sqrt{45}\)
Vì \(\sqrt{49}>\sqrt{45}\)nên \(7>3\sqrt{5}\)
c. Ta có \(\dfrac{1}{3}\sqrt{51}=\dfrac{\sqrt{51}}{3}\), \(\dfrac{1}{5}\sqrt{150}=\sqrt{6}=\dfrac{3\sqrt{6}}{3}=\dfrac{\sqrt{54}}{3}\)
Vì \(\dfrac{\sqrt{51}}{3}< \dfrac{\sqrt{54}}{3}\) nên \(\dfrac{1}{3}\sqrt{51}< \dfrac{1}{5}\sqrt{150}\)
d. Ta có \(\dfrac{1}{2}\sqrt{6}=\dfrac{\sqrt{6}}{2}\), \(6\sqrt{\dfrac{1}{2}}=3\sqrt{2}=\dfrac{6\sqrt{2}}{2}\)
Vì \(\dfrac{\sqrt{6}}{2}< \dfrac{6\sqrt{2}}{2}\Rightarrow\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)