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\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(=\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(5-2\sqrt{6}\right)\)
\(=\sqrt{3}-\sqrt{2}\)
\(=2\sqrt[3]{3}+5\cdot\dfrac{3}{2}-7\cdot4\sqrt[3]{3}+\dfrac{1}{3}\sqrt[3]{3}+5\sqrt[3]{3}\)
\(=-\dfrac{62}{3}\sqrt[3]{3}+\dfrac{15}{2}\)
a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{2}+2\sqrt{3}}\)
= \(\sqrt{18-12}\)
= \(\sqrt{6}\)
b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}.\sqrt{2-2\sqrt{2-\sqrt{2}}}\)
= \(\sqrt{4-4\left(\sqrt{2-\sqrt{2}}\right)^2}\)
= \(\sqrt{4-4.\left(2-4\sqrt{2}+2\right)}\)
= \(\sqrt{4-8+16\sqrt{2}-8}\)
= \(\sqrt{-12+16\sqrt{2}}\)
c)
\(\left(\sqrt{2}-\sqrt{7}\right).\sqrt{9+2\sqrt{14}}\)
= \(\left(\sqrt{2}-\sqrt{7}\right).\left(2+2\sqrt{7}.\sqrt{2}+7\right)\)
= \(\left(\sqrt{2}-\sqrt{7}\right).\left(\sqrt{2}+\sqrt{7}\right)^2\)
= \(\left(4-7\right).\left(\sqrt{2}+\sqrt{7}\right)\)
= \(-3.\left(\sqrt{2}+\sqrt{7}\right)\)
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
Đặt B là tên biểu thức
Với mọi n thuộc N*, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) (*)
Áp dụng (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2^2-\sqrt{3}^2}=\sqrt{4-3}=1\)
a) \(\sqrt{49}+\sqrt{25}-4\cdot0,25\)
\(=7+5-1=11\)
b) \(\sqrt{\frac{1}{9}}\cdot\sqrt{0,81}\cdot\sqrt{0,9}\)
\(=\frac{1}{3}\cdot\frac{9}{10}\cdot\frac{3\sqrt{10}}{10}\)
\(=\frac{9\sqrt{10}}{100}\)
c) \(\sqrt{6,4\cdot2400\cdot0,6}\)
\(=\sqrt{64\cdot36\cdot4}\)
\(=8\cdot6\cdot2=96\)
d) \(\sqrt{26^2-24^2}=\sqrt{\left(26-24\right)\left(26+24\right)}\)
\(=\sqrt{2\cdot50}=\sqrt{100}=10\)