Chứng minh rằng: \(\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\)
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SY
1
30 tháng 6 2019
\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+\frac{2.1}{3}\sqrt{2.3}-\frac{4.1}{2}\sqrt{3.2}\)
\(=\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\)
\(=\sqrt{6}\left(\frac{9}{6}+\frac{4}{6}-\frac{12}{6}\right)=\sqrt{6}.\frac{1}{6}=\frac{\sqrt{6}}{6}\)
Vậy \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}\)
NV
27 tháng 7 2015
có VT \(=\left(\frac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}-\frac{6\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}=\left(\frac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{-3\sqrt{3}}{\sqrt{2}}.\frac{1}{\sqrt{6}}=\frac{-3}{2}\)
dpcm
27 tháng 7 2015
Ta có: \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left\{\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}\right]-\frac{6\sqrt{6}}{3}\right\}\times\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right)\times\frac{1}{\sqrt{6}}\)
\(=\left(-\frac{3\sqrt{6}}{2}\right)\times\frac{1}{\sqrt{6}}\)
\(=\frac{-3}{2}\)(đpcm)
ML
23 tháng 5 2018
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
9 tháng 7 2020
Đặt \(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
Ta có: \(\frac{1}{1+\sqrt{2}}>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)
...
\(\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
Cộng các bất đẳng thức trên lại với nhau, ta được:
\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{81}-1\right)=\frac{1}{2}\cdot\left(9-1\right)=\frac{1}{2}\cdot8=4\)
\(\Leftrightarrow A>4\)(đpcm)
15 tháng 10 2016
Câu trên đề sai
\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\sqrt{2}\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{6}+\sqrt{2}}=1\)
Vậy nó là số nguyên
LM
16 tháng 12 2017
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
______________________________________________
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
\(\sqrt{2+\sqrt{3}}=\sqrt{\frac{1}{2}\left(4+2\sqrt{3}\right)}=\sqrt{\frac{1}{2}}\sqrt{3+2\sqrt{3}+1}=\sqrt{\frac{1}{2}}\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{\frac{1}{2}}.\left(\sqrt{3}+1\right)=\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\left(đpcm\right)\)