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a)\(=\sqrt{\frac{5.5^2}{3^5.2^6}}=\sqrt{\frac{5}{3^5}}.\frac{5}{2^3}=\frac{5\sqrt{5.3^5}}{3^5.2^3}\)
b)\(=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
\(=\frac{2\sqrt{5}}{\sqrt{6}}\)\(=\frac{\sqrt{30}}{3}\)
Câu c ttự
d)\(=\sqrt{2^8.5^2}=2^4.5=80\)
e)\(=\sqrt{\left(\frac{3}{4}\right)^2:\left(\frac{5}{6}\right)^2}=\frac{9}{10}\)
Ta thấy các số trong căn bậc hai đều lớn hơn 0, áp dụng \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)
a) \(\sqrt{7}\cdot\sqrt{63}=\sqrt{7\cdot63}=21\)
b) \(\sqrt{2,5}\cdot\sqrt{30}\cdot\sqrt{48}=\sqrt{2,5\cdot30\cdot48}=60\)
c) \(\sqrt{0,4}\cdot\sqrt{6,4}=\sqrt{0,4\cdot6,4}=1,6\)
d) \(\sqrt{2,7}\cdot\sqrt{5}\cdot\sqrt{1,5}=\sqrt{2,7\cdot5\cdot1,5}=4,5\)
a. \(\sqrt{7}.\sqrt{63}=\sqrt{7.63}=\sqrt{441}=21\)
b.\(\sqrt{2,5}.\sqrt{30}.\sqrt{48}=\sqrt{2,5.30.48}=\sqrt{3600}=60\)
c.\(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{2,56}=1,6\)
d.\(\sqrt{2,7}.\sqrt{5}.\sqrt{1,5}=\sqrt{2,7.5.1,5}=\sqrt{20,25}=4,5\)
\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
a ) \(\sqrt{3+2\sqrt[]{2}}\) - \(\sqrt{2}\)
= \(\sqrt{\left(1+\sqrt{2}\right)^2}\) -\(\sqrt{2}\)
= 1 + \(\sqrt{2}\) - \(\sqrt{2}\)
=1
b) \(\sqrt{16-6\sqrt{7}}\)-\(2\sqrt{7}\)
= \(\sqrt{\left(3-\sqrt{7}\right)^2}\)-\(2\sqrt{7}\)
= 3 - \(\sqrt{7}\)-\(2\sqrt{7}\)
=3 - 3\(\sqrt{7}\)
c )\(\sqrt{30+12\sqrt{6}}\) +\(\sqrt{30-12\sqrt{6}}\)
= \(\sqrt{6\left(5+2\sqrt{6}\right)}\) + \(\sqrt{6\left(5-2\sqrt{6}\right)}\)
=\(\sqrt{6}\) (\(\sqrt{5+2\sqrt{6}}\) + \(\sqrt{5-2\sqrt{6}}\) )
=\(\sqrt{6}\) [\(\sqrt{\left(1+\sqrt{6}\right)^2}\) +\(\sqrt{\left(1-\sqrt{6}\right)^2}\)
=\(\sqrt{6}\) (1 + \(\sqrt{6}\) + \(\sqrt{6}\) -\(1\))
= 2 . 6
=12
d)\(\sqrt{9-4\sqrt{5}}\) -\(\sqrt{5}\)
=\(\sqrt{\left(2-\sqrt{5}\right)}^2\) -\(\sqrt{5}\)
=\(\sqrt{5}\) -\(2\) -\(\sqrt{5}\)
=2
e ) \(\sqrt{\left(-2\right)^6}\) \(+\) \(\sqrt{\left(-3\right)}^4\)
= \(\left|\left(-2\right)^3\right|\) + \(\left|\left(-3\right)^2\right|\)
=8 + 9
=17
a) Ta có: \(\sqrt{0.1}\cdot\sqrt{4000}\)
\(=\sqrt{\frac{1}{10}}\cdot\sqrt{4000}\)
\(=\sqrt{\frac{1}{10}\cdot4000}=\sqrt{400}=20\)
b) Ta có: \(\sqrt{\frac{9}{196}}=\sqrt{\left(\frac{3}{14}\right)^2}\)
\(=\left|\frac{3}{14}\right|\)
\(=\frac{3}{14}\)(Vì \(\frac{3}{14}>0\))
c) Ta có: \(\sqrt{16}\cdot\sqrt{36}-\sqrt{125}:\sqrt{0.01}\)
\(=\sqrt{16\cdot36}-\frac{\sqrt{125}}{\sqrt{\frac{1}{100}}}\)
\(=\sqrt{576}-\sqrt{125:\frac{1}{100}}\)
\(=24-\sqrt{125\cdot100}\)
\(=24-\sqrt{12500}\)
\(=24-50\sqrt{5}\)
d) Ta có: \(\left(\sqrt{112}-\sqrt{63}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(4\sqrt{7}-3\sqrt{3}+\sqrt{7}\right):\sqrt{7}\)
\(=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
e) Ta có: \(\sqrt{2.5}\cdot\sqrt{30}\cdot\sqrt{48}\)
\(=\sqrt{\frac{5}{2}\cdot30\cdot48}=\sqrt{3600}=60\)