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\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
a) ĐS: 2.4.
b) ĐS: 28.
c) HD: Đổi 12,1.360 thành 121.36. ĐS: 66
d) ĐS: 18.
a) \(\sqrt{0,09.64}\)
\(=\sqrt{0,09}.\sqrt{64}\)
\(=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}\)
\(=\sqrt{2^4}.\sqrt{\left(-7\right)^2}\)
\(=2^2.7=4.7=28\)
c) \(\sqrt{12,1.360}\)
\(=\sqrt{121.36}\)
\(=\sqrt{121}.\sqrt{36}\)
\(=11.6=66\)
d) \(\sqrt{2^2.3^4}\)
\(=\sqrt{2^2}.\sqrt{3^4}\)
\(=2.3^2=2.9=18\)
a) = \(\sqrt{10.40}=\sqrt{400}=\sqrt{20^2}=20\)
b) \(=\sqrt{5.45}=\sqrt{5^2.3^2}=\sqrt{15^2}=15\)
a) \(\sqrt{0,09.64}=\sqrt{\left(0,3\right)^2.8^2}=0,3.8=2,4\)
b) \(\sqrt{2^4.\left(-7\right)^2}=\sqrt{\left(2^2\right)^2.\left(-7\right)^2}=2^2.\left|-7\right|=7.4=28\)
c) \(\sqrt{12,1.360}=\sqrt{12,1.10.36}=\sqrt{121.36}=\sqrt{11^2.6^2}=11.6=66\)
d) \(\sqrt{2^2.3^4}=\sqrt{2^2.\left(3^2\right)^2}=2.3^2=9.2=18\)
a) \(\sqrt{0,09\cdot64}=\sqrt{0,09}\cdot\sqrt{64}=0,3\cdot8=2,4\)
b) \(\sqrt{2^4\cdot\left(-7\right)^2}=\sqrt{2^4}\cdot\sqrt{\left(-7\right)^2}=2^2\cdot7=4\cdot7=28\)
c) \(\sqrt{12,1\cdot360}=\sqrt{12,1\cdot10\cdot36}=\sqrt{121\cdot36}=\sqrt{121}\cdot\sqrt{36}=11\cdot6=66\)
d) \(\sqrt{2^2\cdot3^4}=\sqrt{2^2}\cdot\sqrt{3^4}=2\cdot3^2=2\cdot9=18\)
a)\(\sqrt{10}\cdot\sqrt{40}=\sqrt{10\cdot40}=\sqrt{400}=20\)
b) \(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{2\cdot2\cdot81}=\sqrt{4}\cdot\sqrt{81}=2\cdot9=18\)
chỗ \(\sqrt{n}-\sqrt{n+1}\)phải là \(\sqrt{n}+\sqrt{n+1}\)
a, Ta có
\(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{2n+1}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n+1}}< \frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}\)
mà \(\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{2\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng bđt ta có
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
\(=\frac{1}{\left(2\cdot1+1\right)\left(1+\sqrt{2}\right)}+\frac{1}{\left(2\cdot2+1\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2\cdot2011+1\right)\left(\sqrt{2011}-\sqrt{2012}\right)}\)
\(< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\)..
\(=1-\frac{1}{\sqrt{2012}}=\frac{\sqrt{2012}-1}{\sqrt{2012}}=\frac{2011}{\sqrt{2012}\cdot\left(\sqrt{2012}+1\right)}\)
\(=\frac{2011}{2012+\sqrt{2012}}< \frac{2011}{2013}\)
Áp dụng quy tắc khai phương một tích
1: Ta có: \(\sqrt{\frac{1}{5}}\cdot\sqrt{\frac{1}{20}}\cdot3\cdot7\)
\(=\sqrt{\frac{1}{5}}\cdot\sqrt{\frac{1}{20}}\cdot\sqrt{9}\cdot\sqrt{49}\)
\(=\sqrt{\frac{1}{5}\cdot\frac{1}{20}\cdot9\cdot49}\)
\(=\sqrt{\frac{441}{100}}=\frac{\sqrt{441}}{\sqrt{100}}=\frac{21}{10}\)
2: Ta có: \(\sqrt{0,001\cdot360\cdot3^2\cdot\left(-3\right)^2}\)
\(=\sqrt{0,001}\cdot\sqrt{360}\cdot\sqrt{3^{^2}}\cdot\sqrt{\left(-3\right)^2}\)
\(=\sqrt{\frac{1}{100}}\cdot\sqrt{\frac{1}{10}}\cdot\sqrt{6^2}\cdot\sqrt{10}\cdot3\cdot3\)
\(=\frac{1}{10}\cdot6\cdot9\cdot\sqrt{\frac{1}{10}\cdot10}=\frac{54}{10}\cdot1=\frac{27}{5}\)
Áp dụng quy tắc nhân căn thức bậc hai
1: Ta có: \(2\sqrt{2}\left(4\sqrt{8}-\sqrt{32}\right)\)
\(=2\sqrt{2}\cdot4\sqrt{8}-2\sqrt{2}\cdot\sqrt{32}\)
\(=8\cdot\sqrt{16}-2\cdot\sqrt{64}\)
\(=8\cdot4-2\cdot8\)
=32-16=16