Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{1+\sqrt{2}}=\frac{1}{2\sqrt{1}+2\sqrt{2}}+\frac{1}{2\sqrt{1}+2\sqrt{2}}>\frac{1}{2\sqrt{1}+2\sqrt{2}}+\frac{1}{2\sqrt{2}+2\sqrt{3}}\)
\(\Rightarrow\frac{1}{\sqrt{1}+\sqrt{2}}>\frac{1}{2}\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)=\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}\right)\)
Tương tự với các biểu thức còn lại rồi cộng vế với vế ta có:
\(VT>\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\right)=\frac{1}{2}\left(\sqrt{81}-1\right)=4\)
Ta có
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)
Áp dụng vào A ta được
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{80}-\sqrt{79}\)
\(=\sqrt{80}-1>\sqrt{25}-1=4\)
\(\frac{1}{\sqrt{1}+\sqrt{2}}+....\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\) (40 số)
................................................................\(>\frac{40}{10}=4\)
=>đpcm
hc tốt
ko chắc lắm :)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
Tổng quát ta có: Với \(n\inℕ\)ta có:
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\left(n+1\right)-n}{\sqrt{n}+\sqrt{n+1}}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)
Với \(n=2\)\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\)
Với \(n=3\)\(\Rightarrow\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\)
...........................
Với \(n=79\)\(\Rightarrow\frac{1}{\sqrt{79}+\sqrt{80}}=\sqrt{80}-\sqrt{79}\)
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+.....+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{80}-\sqrt{79}\)
\(=\sqrt{80}-\sqrt{2}=\sqrt{40.2}-\sqrt{2}=\sqrt{40}.\sqrt{2}-\sqrt{2}\)
\(=\sqrt{2}.\left(\sqrt{40}-1\right)>\sqrt{2}.\left(\sqrt{36}-1\right)\)
\(=\sqrt{2}.\left(6-1\right)=5\sqrt{2}>4\)( đpcm )
Em thử nhá, ko chắc đâu
1) \(\frac{2}{\sqrt{20}}=\frac{2\sqrt{20}}{20}\) 2) \(\frac{4}{\sqrt{8}}=\frac{4\sqrt{8}}{8}\)
3) \(\frac{2+\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{2}+\sqrt{6}}{2}\) 4) \(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{6-4}=\frac{\sqrt{6}+2}{2}\)
5) \(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=-\left(\sqrt{2}+\sqrt{3}\right)\)
6) \(\frac{9a-b}{3\sqrt{a}-\sqrt{b}}=\frac{\left(9a-b\right)\left(3\sqrt{a}+b\right)}{\left(3\sqrt{a}-\sqrt{b}\right)\left(3\sqrt{a}+\sqrt{b}\right)}=\left(3\sqrt{a}+b\right)\)
7) + 8) em chưa nghĩ ra
ong tth :v
\(\frac{2}{\sqrt{20}}=\frac{\sqrt{4}}{\sqrt{4}.\sqrt{5}}=\frac{1}{\sqrt{5}}\)
\(\frac{4}{\sqrt{8}}=\frac{\sqrt{16}}{\sqrt{8}}=\sqrt{2}\)
\(\frac{2+\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\sqrt{1,5}\)
\(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}=\frac{\sqrt{6}+2}{2}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{3}+\sqrt{2}}{-1}=-\sqrt{3}-\sqrt{2}\)
7: chưa
8: chưa
9:\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Ta có:
\(\frac{1}{\sqrt{1}+\sqrt{2}}>\frac{1}{\sqrt{2}+\sqrt{3}};\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{\sqrt{4}+\sqrt{5}};...;\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{80}+\sqrt{81}}\)
Do đó \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}\right)\)\(>\frac{1}{2}\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(=\frac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{80}-\sqrt{79}+\sqrt{81}-\sqrt{80}\right)\)
\(=\frac{1}{2}\left(-\sqrt{1}+\sqrt{81}\right)=\frac{1}{2}\left(-1+9\right)=4\)
Suy ra đpcm.
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{79}}\)
Suy ra
\(2A=2\left(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\right)\)
\(=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+....+\left(\sqrt{80}-\sqrt{79}\right)+\left(\sqrt{81}-\sqrt{79}\right)\)
\(=\sqrt{81}-1=9-1=8\Rightarrow2A>8\Leftrightarrow A>8\)( Đpcm)