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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: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
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 : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
a)
<=> \(\dfrac{7}{4\cdot\sqrt{3}}và\dfrac{9}{4\cdot\sqrt{5}}\)
<=> \(\dfrac{7\cdot\sqrt{5}}{4\cdot\sqrt{15}}và\dfrac{9\cdot\sqrt{3}}{4\cdot\sqrt{15}}\)
<=>\(\sqrt{245}và\sqrt{243}\)
<=> \(\sqrt{245}>\sqrt{243}\)
=> \(\dfrac{7}{2}\cdot\sqrt{\dfrac{1}{12}}=\dfrac{9}{4}\cdot\sqrt{\dfrac{1}{5}}\)
a)
\(\dfrac{7}{2}\sqrt{\dfrac{1}{12}}=\dfrac{7}{2}\sqrt{\dfrac{12}{12^2}}=\dfrac{7}{2}.\dfrac{\sqrt{12}}{\sqrt{12^2}}=\dfrac{7}{2}.\dfrac{\sqrt{3.4}}{12}=\dfrac{7.2.\sqrt{3}}{2.12}=\dfrac{7\sqrt{3}}{12}=\dfrac{7\sqrt{3}.5}{12.5}=\dfrac{35\sqrt{3}}{60}\)
\(\dfrac{9}{4}\sqrt{\dfrac{1}{5}}=\dfrac{9}{4}\sqrt{\dfrac{5}{5^2}}=\dfrac{9}{4}.\dfrac{\sqrt{5}}{\sqrt{5^2}}=\dfrac{9.\sqrt{5}}{4.5}=\dfrac{9\sqrt{5}}{20}=\dfrac{9\sqrt{5}.3}{20.3}=\dfrac{27\sqrt{5}}{60}\)Ta có \(3675>3645\Leftrightarrow\sqrt{3675}>\sqrt{3645}\Leftrightarrow\sqrt{1225.3}>\sqrt{729.5}\Leftrightarrow35\sqrt{3}>27\sqrt{5}\Leftrightarrow\dfrac{35\sqrt{3}}{60}>\dfrac{27\sqrt{5}}{60}\)
Vậy \(\dfrac{7}{2}\sqrt{\dfrac{1}{12}}>\dfrac{9}{4}\sqrt{\dfrac{1}{5}}\)
b)
\(\sqrt{\dfrac{4}{27}}=\sqrt{\dfrac{4.3}{27.3}}=\dfrac{\sqrt{4.3}}{\sqrt{81}}=\dfrac{2\sqrt{3}}{9}=\dfrac{2\sqrt{3}.26}{9.26}=\dfrac{52\sqrt{3}}{234}\)
\(\sqrt{\dfrac{3}{26}}=\sqrt{\dfrac{3.26}{26^2}}=\dfrac{\sqrt{3.26}}{\sqrt{26^2}}=\dfrac{\sqrt{78}}{26}=\dfrac{9.\sqrt{78}}{26.9}=\dfrac{9\sqrt{78}}{234}\)
Ta có \(8112>6318\Leftrightarrow\sqrt{8112}>\sqrt{6318}\Leftrightarrow\sqrt{2704.3}>\sqrt{81.78}\Leftrightarrow52\sqrt{3}>9\sqrt{78}\Leftrightarrow\dfrac{52\sqrt{3}}{234}>\dfrac{9\sqrt{78}}{234}\)
Vậy \(\sqrt{\dfrac{4}{27}}>\sqrt{\dfrac{3}{26}}\)