cmr \(\sqrt{5}-\sqrt{3}>\sqrt{3}-\sqrt{2}\)
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)
Đặt \(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}=x>0\)
\(\Rightarrow x^3=14-3\left(\sqrt[3]{5\sqrt[]{2}+7}-\sqrt[3]{5\sqrt[]{2}-7}\right)\sqrt[3]{\left(5\sqrt[]{2}+7\right)\left(5\sqrt[]{2}-7\right)}\)
\(\Rightarrow x^3=14-3x.\sqrt[3]{\left(5\sqrt[]{2}\right)^2-7^2}\)
\(\Rightarrow x^3=14-3x\)
\(\Rightarrow x^3+3x-14=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)=0\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{8}+\sqrt{10}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{4}+\sqrt{2}.\sqrt{5}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)\left(1+\sqrt{2}\right)}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)
\(=1+\sqrt{2}\)
⇒ ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(\Rightarrow n^2=\left(3+\sqrt{5+2\sqrt{3}}\right)+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}+\left(3-\sqrt{5+2\sqrt{3}}\right)\)
\(=6+2\sqrt{9-\left(5+2\sqrt{3}\right)}\)
\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=6+2\sqrt{3}-2=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow n=\sqrt{3}+1\)
\(\Rightarrow n^2-2n-2=\left(4+2\sqrt{3}\right)-2\left(\sqrt{3}+1\right)-2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\sqrt{2\sqrt{3\sqrt{4....\sqrt{2017}}}}\)
< \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2016\sqrt{2018}}}}}\)
\(=\sqrt{2\sqrt{3\sqrt{4...\sqrt{2017^2-1}}}}\)
< \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2015.2017}}}}\)
.......................................................................
< \(\sqrt{2.4}< \sqrt{9}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài đúng: \(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)
Hoặc: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=1\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)