Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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.\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
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 \(\dfrac{\sqrt{6}}{2}< \dfrac{6\sqrt{2}}{2}\Rightarrow\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b: \(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5\sqrt{6}}{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
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: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
\(A=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}=\dfrac{5-2\sqrt{6}}{25-24}-\dfrac{5+2\sqrt{6}}{25-24}=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)
-
\(B=\dfrac{1}{\sqrt{3}+2}-\dfrac{1}{\sqrt{3}-2}=\dfrac{\sqrt{3}-2}{3-4}-\dfrac{\sqrt{3}+2}{3-4}=-\sqrt{3}+2+\sqrt{3}+2=4\)
-
\(C=\dfrac{3}{\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\sqrt{3}+\sqrt{3}\left(\sqrt{3}-1\right)=\sqrt{3}+3-\sqrt{3}=3\)
-
\(D=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}==\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{1}-\dfrac{2+\sqrt{3}}{1}=5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}-2+\sqrt{3}=-2\)
a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)