Tính:
a, \(\sqrt{20}\)*\(\sqrt{72}\)*\(\sqrt{4,9}\)
b,\(\sqrt{\frac{999}{111}}\)
c,\(\sqrt{146,5^2-109,5^2+27256}\)
d,\(\sqrt{\frac{149^2-76^2}{457^2-348^2}}\)
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31 tháng 5 2017
a) \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}\)
\(=\sqrt{\frac{11849}{164}}=\sqrt{72,25}=8,5\)
b)\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\) \(=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}\)
c)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\) \(=\sqrt{2^2+3+2.2.\sqrt{3}}-\sqrt{2^2+3-2.2.\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
8 tháng 8 2016
1) \(\frac{\sqrt{165^2-124^2}}{164}=\frac{\sqrt{\left(165-124\right)\left(165+124\right)}}{164}=\frac{\sqrt{41}\cdot\sqrt{289}}{164}=\frac{\sqrt{41}\cdot17}{164}=\frac{17}{4\sqrt{41}}\)
2) \(\frac{\sqrt{149^2-76^2}}{\sqrt{457^2-384^2}}=\frac{\sqrt{\left(149+76\right)\left(149-76\right)}}{\sqrt{\left(457+384\right)\left(457-384\right)}}=\frac{\sqrt{225}\cdot\sqrt{73}}{\sqrt{841}\cdot\sqrt{73}}=\frac{25}{29}\)
30 tháng 5 2017
a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)
b) tương tự ý a
c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)
\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=14-2\sqrt{49-48}\)
\(=14-2.1=12\)
\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)
18 tháng 7 2017
\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)
LD
21 tháng 9 2017
a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)
=\(\sqrt{3,6.10}=\sqrt{36}=6\)
b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)
=\(\sqrt{3,6.40}=\sqrt{144}=12\)
c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)
d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)
19 tháng 5 2022
a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
NT
1
14 tháng 8 2016
a) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
\(=\sqrt{91.144-1440}=\sqrt{144\left(91-10\right)}=\sqrt{12^2.9^2}=12.9=108\)
b) \(\sqrt{146,5^2-109,5^2+27.256}=\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
\(=\sqrt{37.256+27.256}=\sqrt{256\left(37+27\right)}=\sqrt{256.64}=\sqrt{16^2.8^2}=16.8=128\)
9 tháng 6 2017
2)
- \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003\times2005}\)
\(=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}=4008+2\sqrt{2004^2-1}\)
- \(\left(\sqrt{2004}+\sqrt{2004}\right)^2=2004+2004+2\sqrt{2004\times2004}\)
\(=4008+2\sqrt{2004^2}\)
Ta có \(2004^2>2004^2-1\Rightarrow\sqrt{2004^2}>\sqrt{2004^2-1}\Rightarrow4008+2\sqrt{2004^2}>4008+2\sqrt{2004^2-1}\)
Vậy \(2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)
a. =\(\sqrt{20.72.4,9}=\sqrt{2.72.49}=\sqrt{144.49}=12.7=84\)
b. \(\sqrt{\frac{999}{111}}=\sqrt{9}=3\)
c. = \(\sqrt{9472+27256}=\sqrt{36728}\approx191,645\)
d. = \(\sqrt{\frac{\left(149+76\right)\left(149-76\right)}{\left(457+348\right)\left(457-348\right)}}=\sqrt{\frac{225.73}{805.109}}=\sqrt{\frac{3285}{17549}}\approx136,817\)