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![](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)
a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)
\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)
Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)
b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)
\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)
Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)
mà 80>75
nên \(4\sqrt{5}>5\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a^3 = 5\(\sqrt{2}\) b^3= 5\(\sqrt[3]{2}\).\(\sqrt{5\sqrt[3]{2}}\)
ta co \(\sqrt{5\sqrt[3]{2}}\)>2 >\(\sqrt{2}\)
=> b^3 >a^3 => b>a
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\sqrt{a^2}=\left|a\right|\)
\(\sqrt[3]{a^3}=a\)
b: \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)
Số giá trị nguyên thỏa mãn điều kiện là:
\(\left(2+4\right)+1=7\)
\(a,\sqrt{3}và1,7\left(3\right)=1,73205...và1,7\left(3\right)\\ \Rightarrow1,73205>1,7\left(3\right)\\ \Rightarrow\sqrt{3}>1,7\left(3\right).\\ b,-2,236và-\sqrt{5}=-2,236và-2,23606...\\ \Rightarrow-2,236>-2,23606\\ \Rightarrow-2,236>-\sqrt{5} \)
a) sai