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\(a,2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}.\sqrt{\dfrac{75}{16}}\)
\(\Leftrightarrow2.\dfrac{\sqrt{27}}{2}-\sqrt{\dfrac{48}{3}}-\dfrac{2}{5}.\dfrac{\sqrt{75}}{4}\)
\(\Leftrightarrow\sqrt{27}-\dfrac{4\sqrt{3}}{3}-\dfrac{1}{5}.\dfrac{5\sqrt{3}}{2}\)
\(\Leftrightarrow3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\dfrac{7\sqrt{3}}{6}\)
\(b,\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right).\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\Leftrightarrow\)\(\left[1+\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{-4}\right].\left[\dfrac{\left(5+\sqrt{5}\right).\left(1-\sqrt{5}\right)}{-4}+1\right]\)
\(\Leftrightarrow\)\(\left(1+\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{-4}\right).\left(\dfrac{5-5\sqrt{5}+\sqrt{5}-5}{-4}+1\right)\)
\(\Leftrightarrow\)\(\left(1+\dfrac{4\sqrt{5}}{-4}\right)\left(\dfrac{-4\sqrt{5}}{-4}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right).\left(1+\sqrt{5}\right)\)
<=> 1-5
=-4
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 :
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 : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)
b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)
c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)
d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)
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 \(\sqrt{\dfrac{4}{27}}>\sqrt{\dfrac{3}{26}}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
\(\dfrac{2}{3}\sqrt{9u-9}-\dfrac{1}{4}\sqrt{16u-16}+27\sqrt{\dfrac{u-1}{81}}=4\left(dk:u\ge1\right)\)
\(\Leftrightarrow\dfrac{2}{3}\sqrt{9\left(u-1\right)}-\dfrac{1}{4}\sqrt{16\left(u-1\right)}+27\dfrac{\sqrt{u-1}}{\sqrt{81}}=4\)
\(\Leftrightarrow2\sqrt{u-1}-\sqrt{u-1}+3\sqrt{u-1}=4\\ \Leftrightarrow\sqrt{u-1}.\left(2-1+3\right)=4\\ \Leftrightarrow4\sqrt{u-1}=4\\ \Leftrightarrow\sqrt{u-1}=1\\ \Leftrightarrow u-1=1\\ \Leftrightarrow u=2\left(tm\right)\)
Vậy \(S=\left\{2\right\}\)