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a, \(3\sqrt{3}\) >\(2\sqrt{3}\) =>\(3\sqrt{3}\) >\(\sqrt{12}\)
b,có \(3\sqrt{5}=\sqrt{45}\) <\(\sqrt{49}=7\) =>7 >\(3\sqrt{5}\)
c,\(\sqrt{\dfrac{51}{9}}\) <\(\sqrt{6}\) => \(\dfrac{1}{3}\sqrt{51}\) <\(\dfrac{1}{5}\sqrt{150}\)
d.\(\dfrac{1}{2}\sqrt{6}< 6\sqrt{\dfrac{1}{2}}\)
~ ~ ~
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\)
\(=\sqrt{\dfrac{37}{4}-\sqrt{\left(3\sqrt{5}+2\right)^2}}\)
\(=\sqrt{\dfrac{29}{4}-3\sqrt{5}}\)
\(=\sqrt{\dfrac{29-12\sqrt{5}}{4}}\)
\(=\sqrt{\dfrac{\left(2\sqrt{5}-3\right)^2}{4}}\)
\(=\dfrac{\sqrt{5}}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}\left(\sqrt{5}-\dfrac{3}{2}\right)\)
\(>\sqrt{5}-\dfrac{3}{2}=B\)
~ ~ ~
\(C=\dfrac{16\sqrt{36}-20\sqrt{48}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-80\sqrt{3}+10\sqrt{3}}{\sqrt{12}}\)
\(=\dfrac{96-70\sqrt{3}}{2\sqrt{3}}\)
\(=16\sqrt{3}-35\)
\(>16\sqrt{3}-36=B\)
~ ~ ~
a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)
=> A2=8+2\(\sqrt{3}\)
B=\(\sqrt{3}+1\)=> B2=10+2\(\sqrt{3}\)
=>A>B
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
Võ Đông Anh Tuấn
Áp dụng \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
a)
\(7=\sqrt{49}\\ 3\sqrt{5}=\sqrt{9}\cdot\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\\ \text{Vì }\sqrt{49}>\sqrt{45}\text{ nên }7>3\sqrt{5}\)
Vậy \(7>3\sqrt{5}\)
b)
\(2\sqrt{7}+3=\sqrt{4}\cdot\sqrt{7}+3=\sqrt{4\cdot7}+3=\sqrt{28}+3\\ \sqrt{28}+3>\sqrt{25}+3=5+3=8\)
Vậy \(8< 2\sqrt{7}+3\)
c)
\(3\sqrt{6}=\sqrt{9}\cdot\sqrt{6}=\sqrt{9\cdot6}=\sqrt{54}\\ 2\sqrt{15}=\sqrt{4}\cdot\sqrt{15}=\sqrt{4\cdot15}=\sqrt{60}\\ \text{Vì } \sqrt{54}< \sqrt{60}\text{nên }3\sqrt{6}< 2\sqrt{15}\)
Vậy \(3\sqrt{6}< 2\sqrt{15}\)
a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)
b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)
a: \(2\sqrt{5}=\sqrt{20}\)
\(5=\sqrt{25}\)
mà 20<25
nên \(2\sqrt{5}< 5\)
b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)
\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)
mà 16<9
nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)