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Giả sử \(8>\sqrt{15}+\sqrt{17}\)
\(\Leftrightarrow64>32+2\sqrt{15×17}\)
\(\Leftrightarrow16>\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\left(dung\right)\)
Vậy \(8>\sqrt{15}+\sqrt{17}\)
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\(8^2=64=32+2\sqrt{16^2}\)
\(\left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{15.17}=32+2\sqrt{\left(16-1\right)\left(16+1\right)}\)
\(=32+2\sqrt{16^2-1}\)
\(< =>8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\)
\(8>\sqrt{15}+\sqrt{17}\)
\(\left(\sqrt{2019}+\sqrt{2021}\right)^2=4040+2\sqrt{2019.2021}\)
\(=4040+2\sqrt{\left(2020-1\right)\left(2020+1\right)}=4040+2\sqrt{2020^2-1}\)
\(\left(2\sqrt{2020}\right)^2=8080=4040+2\sqrt{2020^2}\)
\(< =>\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
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8 lớn hơn \(\sqrt{15}\)+\(\sqrt{17}\)
vì \(\sqrt{15}\)+\(\sqrt{17}\)=7,997,,
![](https://rs.olm.vn/images/avt/0.png?1311)
Giả sử \(8< \sqrt{15}+\sqrt{17}\)
\(\Leftrightarrow64< 15+2\sqrt{15.17}+17\)(Bình phương hai vế)
\(\Leftrightarrow32< 2\sqrt{15.17}\)
\(\Leftrightarrow16< \sqrt{15.17}\)
\(\Leftrightarrow16< \sqrt{\left(16-1\right)\left(16+1\right)}\)
\(\Leftrightarrow\sqrt{16^2}< \sqrt{16^2-1}\)
\(\Leftrightarrow16^2< 16^2-1\)(vô lí)
Chứng minh tương tự điều giả sử \(8=\sqrt{15}+\sqrt{17}\)
Vậy \(8>\sqrt{15}+\sqrt{17}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)
Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)
\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)
\(\Rightarrow A< B\)
\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)
\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)
mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)
nên A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\sqrt{15}< \sqrt{16}=4$
$\sqrt{17}< \sqrt{25}=5$
$\Rightarrow \sqrt{15}+\sqrt{17}< 9< 16$
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(\sqrt{17}>\sqrt{15}\)
\(\Rightarrow\sqrt{33}-\sqrt{15}>\sqrt{33}-\sqrt{17}\)
\(\sqrt{36}-\sqrt{15}>\sqrt{33}-\sqrt{15}\Leftrightarrow6-\sqrt{15}>\sqrt{33}-\sqrt{15}\)
Vậy \(6-\sqrt{15}>\sqrt{33}-\sqrt{17}\)vì \(\sqrt{33}-\sqrt{15}>\sqrt{33}-\sqrt{17}\)
\(8^2=64=32+32\\ \left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{255}\)
\(32^2=1024>1020=\left(2\sqrt{255}\right)^2\\ \Rightarrow64>32+2\sqrt{255}\\ \Rightarrow8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\\ \Leftrightarrow8>\sqrt{15}+\sqrt{17}\\ \Leftrightarrow-8< -\left(\sqrt{15}+\sqrt{17}\right)\)
Lời giải:
\((\sqrt{15}+\sqrt{17})^2=32+2\sqrt{15.17}=32+2\sqrt{(16-1)(16+1)}\)
\(=32+2\sqrt{16^2-1}< 32+2\sqrt{16^2}=64\)
\(\Rightarrow \sqrt{15}+\sqrt{17}< 8\)
\(\Rightarrow -(\sqrt{15}+\sqrt{17})> -8\)