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\(A=\sqrt{6+2\sqrt{5}}-\sqrt{5}=\sqrt{5}+1-\sqrt{5}=1\)
\(B=\sqrt[3]{7+5\sqrt{2}}-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)
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
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a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
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Lời giải:
\(\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{11}}> \frac{1}{\sqrt{4}}+\frac{1}{\sqrt{9}}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
căn 24< căn 25 =5 :
căn 45<căn 49 =7
=> căn 24+ căn 45 < căn 25+ căn 49 =5+7=12
a) \(2\sqrt{3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{18}\)
Vì 12<18 => \(\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) \(12=5+7=\sqrt{25}+\sqrt{49}>\sqrt{24}+\sqrt{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
Có: \(2\sqrt{3}=\sqrt{12}>\sqrt{9}=3\)
\(3\sqrt{2}=\sqrt{18}>\sqrt{16}=4\)
Vì thế: \(2\sqrt{3}+3\sqrt{2}>7\)