So sánh hai số A,B biết \(A=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(B=\left(\frac{1}{\sqrt{\sqrt{5}+2}}+\sqrt{\sqrt{5}+2}\right)\cdot\frac{1}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
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Y
13 tháng 6 2019
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 : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
Y
13 tháng 6 2019
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 :))
23 tháng 7 2020
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
PT
23 tháng 7 2020
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4
S
13 tháng 6 2019
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:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
15 tháng 7 2016
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
15 tháng 7 2016
\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
25 tháng 7 2022
a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(=\dfrac{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}}{\sqrt{2}}\)
\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
10 tháng 7 2019
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
10 tháng 7 2019
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
Ta có: \(A=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2^2-\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-\left|2+\sqrt{3}\right|}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-\left(2+\sqrt{3}\right)}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2^2-\left(\sqrt{3}\right)^2}\)
\(=\sqrt{4-3}\)(Vì \(3>0\))
\(=1\)
Ta có: \(B=\left(\frac{1}{\sqrt{\sqrt{5}+2}}+\sqrt{\sqrt{5}+2}\right)\cdot\frac{1}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\left(\frac{1}{\sqrt{\sqrt{5}+2}}+\frac{\sqrt{5}+2}{\sqrt{\sqrt{5}+2}}\right)\cdot\frac{\sqrt{\sqrt{5}-1}}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{3+\sqrt{5}}{\sqrt{\sqrt{5}+2}}\cdot\frac{\sqrt{\sqrt{5}-1}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\left(3+\sqrt{5}\right)}{2}\cdot\sqrt{\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}}-\left|\sqrt{2}-1\right|\)
\(=\frac{3+\sqrt{5}}{2}\cdot\sqrt{\frac{7-3\sqrt{5}}{5-4}}-\left(\sqrt{2}-1\right)\)(Vì \(\sqrt{2}>1\))
\(=\frac{6+2\sqrt{5}}{4}\cdot\sqrt{7-3\sqrt{5}}-\sqrt{2}+1\)
\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\sqrt{14-6\sqrt{5}}}{4\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\left(3-\sqrt{5}\right)}{4\sqrt{2}}-\frac{4\sqrt{2}\cdot\left(\sqrt{2}-1\right)}{4\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\left(6-2\sqrt{5}\right)}{8\sqrt{2}}-\frac{8\sqrt{2}\cdot\left(\sqrt{2}-1\right)}{8\sqrt{2}}\)
\(=\frac{16-16+8\sqrt{2}}{8\sqrt{2}}\)
=1
Vậy: A=B