so sánh \(\sqrt[5]{-0.3 } \) và \(\sqrt[3]{-0.4}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt:
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)
\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)
\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)
\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)
Mà: \(4\sqrt{5}>1\)
Nên: \(A^2< B^2\)
\(\Rightarrow A< B\)
Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
=>A^2=(căn 10)^2=10=9+1
Đặt B=2+căn 5
=>B^2=(2+căn 5)^2=9+4căn 5
1<4căn 5
=>9+1<9+4căn 5
=>A^2<B^2
=>A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)
Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)
Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)
Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)
Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 3 + √2 < 3 + 2 = 5
Mặt khác 3 + √2 > 1
Do đó: √3+√23+2 < √5
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)
\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
Do đó: A=B
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}=\left|\sqrt{5}+1\right|-\sqrt{5}=1\)
\(\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}\right)^3+1^3+3.2+3\sqrt{2}}-\sqrt{2}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
--> Bằng nhau