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a,\(\dfrac{1}{4}\sqrt{8}=\dfrac{1}{\sqrt{2}}\)
\(\dfrac{2}{3}\sqrt{12}=\dfrac{4}{\sqrt{3}}\)
=> \(\dfrac{1}{4}\sqrt{8}< \dfrac{2}{3}\sqrt{12}\)
Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
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
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, \(3\sqrt{3}\) >\(2\sqrt{3}\) =>\(3\sqrt{3}\) >\(\sqrt{12}\)
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: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
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: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
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: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
So Sánh
a.\(\dfrac{1}{4}\sqrt{8}\) và \(\dfrac{2}{3}\sqrt{12}\)
Có:\(\dfrac{1}{4}\sqrt{8}\) và \(\dfrac{2}{3}\sqrt{12}\)
= \(\dfrac{1}{4}.2\sqrt{2}\) và \(\dfrac{2}{3}.2\sqrt{3}\)
=\(\dfrac{\sqrt{2}}{2}\)và \(\dfrac{4\sqrt{3}}{3}\)
=> \(\dfrac{1}{4}\sqrt{8}< \dfrac{2}{3}\sqrt{12}\)
b. \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}\)và \(6\sqrt{\dfrac{1}{35}}\)
Có \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}\) và \(6\sqrt{\dfrac{1}{35}}\)
=\(\dfrac{5}{2}.\dfrac{\sqrt{6}}{6}\) và \(6.\dfrac{\sqrt{35}}{35}\)
=\(\dfrac{5\sqrt{6}}{12}\) và \(\dfrac{6\sqrt{35}}{35}\)
=> \(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{35}}\)
c. \(\dfrac{1}{6}\sqrt{18}\) và \(\dfrac{1}{2}\sqrt{2}\)
=\(\dfrac{1}{6}.3\sqrt{2}\) và \(\dfrac{1}{2}\sqrt{2}\)
=\(\dfrac{\sqrt{2}}{2}\) và \(\dfrac{\sqrt{2}}{2}\)
=> \(\dfrac{1}{6}\sqrt{18}=\dfrac{1}{2}\sqrt{2}\)