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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/ \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}=7-4\sqrt{3}+7+4\sqrt{3}=14\)
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 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
1. với a=2,5 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|2.5\right|=2.5\)
với a=0,3 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|0,3\right|=0,3\)
với a=-0,1 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|-0,1\right|=0,1\)
Bài 4:
a: \(=\sqrt{\dfrac{10.8}{0.3}}=\sqrt{36}=6\)
b: \(=\sqrt{\dfrac{7}{175}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
c: \(=\sqrt{\dfrac{2.84}{0.71}}=2\)
d: \(=\sqrt{\dfrac{625}{144}}=\dfrac{25}{12}\)
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: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
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, \(3\sqrt{3}\) >\(2\sqrt{3}\) =>\(3\sqrt{3}\) >\(\sqrt{12}\)
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: \(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
Vì 112 < 117 => \(\sqrt{112}< \sqrt{117}\) => \(4\sqrt{7}< 3\sqrt{13}\)
b) Ta có: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{21}{4}}=\sqrt{\dfrac{147}{28}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}=\sqrt{\dfrac{114}{28}}\)
Vì: \(\dfrac{147}{28}>\dfrac{114}{28}\Rightarrow\sqrt{\dfrac{147}{28}}>\sqrt{\dfrac{114}{28}}\Rightarrow\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)