![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,2\sqrt{3}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}=\sqrt{75}>\sqrt{74}\\ 2,\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\\ \left(2\sqrt{2}+\sqrt{6}\right)^2=14+4\sqrt{3}\\ 6\sqrt{5}=\sqrt{180}>\sqrt{48}=4\sqrt{3}\\ \Rightarrow3+\sqrt{5}>2\sqrt{2}+\sqrt{6}\\ 3,3\sqrt{3}+\sqrt{26}+1>3\sqrt{3}+\sqrt{3}\left(\sqrt{26}>\sqrt{3}\right)\\ \Rightarrow\sqrt{27}+\sqrt{26}+1>4\sqrt{3}=\sqrt{48}\\ 4,\dfrac{1}{\sqrt{15}+\sqrt{14}}< \dfrac{1}{\sqrt{14}+\sqrt{13}}\left(\sqrt{15}+\sqrt{14}>\sqrt{14}+\sqrt{13}\right)\\ \Rightarrow\sqrt{15}-\sqrt{14}< \sqrt{14}-\sqrt{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a : Cộng 2 vế cho 6 ta được :
\(7+6......7+\sqrt{37}\)
Mà : \(6=\sqrt{36}< \sqrt{37}\)
\(\Rightarrow7+6< \sqrt{37}+1\)
\(\Rightarrow7< \sqrt{37}+1\)
Cách khác của câu a.
Ta có : \(\sqrt{37}>\sqrt{36}=6\)
\(\Rightarrow\sqrt{37}+1>6+1=7\)
Vậy \(\sqrt{37}+1>7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Ta có: \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+2\sqrt{6}+3=5+2\sqrt{6}\)
Vì \(2^2=4< 5\Rightarrow2^2< 5+2\sqrt{6}\Rightarrow2^2< \left(\sqrt{2}+\sqrt{3}\right)^2\)
Do đó : \(\sqrt{2}+\sqrt{3}>2\)
b/ Ta có: \((\sqrt{24}+\sqrt{45})^2=24+45+2\sqrt{4.6.9.5}=69+12\sqrt{30}\)
\(12^2=144=69+75\)
Lại có: \(\left(12\sqrt{30}\right)^2=144.30=4320\)
\(75^2=5625\)
Vì \(4320< 5625\Rightarrow12\sqrt{30}< 75\Rightarrow12\sqrt{30}+69< 75+69\Rightarrow\left(\sqrt{24}+\sqrt{45}\right)^2< 12^2\Rightarrow\sqrt{24}+\sqrt{45}< 12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo bà ra ta có :
\(f\left(2\sqrt{3}\right)=\left(m+1\right)x-2=\left(m+1\right)\left(2\sqrt{3}\right)-2\)
\(=\sqrt{12}\left(m+1\right)-2\)
\(f\left(3\sqrt{2}\right)=\left(m+1\right)x-2=\left(m+1\right)3\sqrt{2}-2\)
\(=\sqrt{18}\left(m+1\right)-2\)
vì 12 < 18 => \(\sqrt{12}< \sqrt{18}\)
hay \(f\left(2\sqrt{3}\right)< f\left(3\sqrt{2}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{2004}}\)
Xét số hạng tổng quát: \(\frac{1}{\sqrt{n}}\) ta có:
\(\frac{1}{\sqrt{n}}=\frac{2}{2\sqrt{n}}> \frac{2}{\sqrt{n}+\sqrt{n+1}}=\frac{2(\sqrt{n+1}-\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}=2(\sqrt{n+1}-\sqrt{n})\)
Do đó:
\(\frac{1}{\sqrt{1}}> 2(\sqrt{2}-\sqrt{1})\)
\(\frac{1}{\sqrt{2}}> 2(\sqrt{3}-\sqrt{2})\)
\(\frac{1}{\sqrt{3}}> 2(\sqrt{4}-\sqrt{3})\)
............
\(\frac{1}{\sqrt{2004}}> 2(\sqrt{2005}-\sqrt{2004})\)
Cộng theo vế:
$A>2(\sqrt{2005}-1)>86$
Vậy..........
1/2√26 < 1/3√63
Có cần cách làm hơm