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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 \(3\sqrt{6}< 2\sqrt{15}\)
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) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
\(1)\) 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 ~
* \(4\)và \(1+2\sqrt{2}\)
Ta có \(3=\sqrt{9}\)
\(2\sqrt{2}=\sqrt{2^2.2}=\sqrt{8}\)
Ta lại có \(8< 9\Leftrightarrow\sqrt{8}< \sqrt{9}\)
Hay \(2\sqrt{2}< 3\)\(\Leftrightarrow1+2\sqrt{2}< 1+3\Leftrightarrow1+2\sqrt{2}< 4\)
a) Ta có:
\(6\sqrt{5}=\sqrt{5\cdot36}=\sqrt{180}\)
\(5\sqrt{6}=\sqrt{6\cdot25}=\sqrt{200}\)
Mà \(\sqrt{180}< \sqrt{200}\)
Vậy: \(6\sqrt{5}< 5\sqrt{6}\)
x) Ta có: \(\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\)
Công hai vế của BĐT cho 3:
Suy ra: \(\sqrt{8}+3< 3+3=6\)
Vậy: \(\sqrt{8}+3< 6\)
b) Ta có:
\(\sqrt{2\sqrt{3}}=\sqrt[4]{12}\)
Tương tự: \(\sqrt{3\sqrt{2}}=\sqrt[4]{18}\)
Mà \(\sqrt[4]{18}>\sqrt[4]{12}\)
Vậy.....
d) Ta có:
\(2\sqrt{5}-5=\sqrt{5}+\sqrt{5}-5=\left(\sqrt{5}-2\right)+\left(\sqrt{5}-3\right)>\sqrt{5}-3\)
Vậy ......
e) Ta có:
\(\sqrt{2}-2=\frac{3\sqrt{2}-6}{3}\)
\(\sqrt{3}-3=\frac{2\sqrt{3}-6}{2}\)
Mà \(3\sqrt{2}>2\sqrt{3}\)
Vậy .....
f) ........... Đang thinking
a: \(\left(\sqrt{3}+\sqrt{5}\right)^2=8+\sqrt{60}\)
\(\left(\sqrt{17}\right)^2=17=8+\sqrt{81}\)
mà 60<81
nên \(3+\sqrt{5}< \sqrt{17}\)
c: \(\left(\sqrt{2004}+\sqrt{2006}\right)^2=4010+2\cdot\sqrt{2005^2-1}\)
\(\left(2\cdot\sqrt{2005}\right)^2=8020=4010+2\cdot\sqrt{2005^2}\)
mà \(2005^2-1< 2005^2\)
nên \(\sqrt{2004}+\sqrt{2006}< 2\sqrt{2005}\)
d: \(\left(\sqrt{5}+2\right)^2=9+4\sqrt{5}=9+\sqrt{80}\)
\(\left(\sqrt{3}+\sqrt{6}\right)^2=9+2\cdot\sqrt{3\cdot6}=9+\sqrt{72}\)
mà 80>72
nên \(\sqrt{5}+2>\sqrt{3}+\sqrt{6}\)
\(a\)
\(\sqrt{7}+\sqrt{15}\)
\(=\sqrt{7+15}\)
\(=4,69\)
\(4,69< 7\)
\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)
\(b\)
\(\sqrt{7}+\sqrt{15}+1\)
\(=\sqrt{7+15}+1\)
\(=4,69+1\)
\(=5,69\)
\(\sqrt{45}\)
\(=6,7\)
\(5,69< 6,7\)
\(\Rightarrow\)\(\sqrt{7}+\sqrt{15}+1\)\(< \)\(\sqrt{45}\)
\(c\)
\(\frac{23-2\sqrt{19}}{3}\)
\(=\frac{22.4,53}{3}\)
\(=\frac{95,7}{3}\)
\(=31,9\)
\(\sqrt{27}\)
\(=5,19\)
\(31,9>5,19\)
\(\text{}\Rightarrow\text{}\text{}\)\(\frac{23-2\sqrt{19}}{3}\)\(>\sqrt{27}\)
\(d\)
\(\sqrt{3\sqrt{2}}\)
\(=\sqrt{3.1,41}\)
\(=\sqrt{4,23}\)
\(=2,05\)
\(\sqrt{2\sqrt{3}}\)
\(=\sqrt{2.1,73}\)
\(=\sqrt{3,46}\)
\(=1,86\)
\(2,05>1,86\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
\(Học \) \(Tốt !!!\)
a) Ta có : \(\sqrt{7}< \sqrt{9}=3;\sqrt{15}< \sqrt{16}=4\)
Do đó : \(\sqrt{7}+\sqrt{15}< 3+4=7\)
b) Ta có : \(\sqrt{17}>\sqrt{16}=4;\sqrt{5}>\sqrt{4}=2\)
\(\Rightarrow\sqrt{17}+\sqrt{5}+1>4+2+1=7\)
Lại có : \(\sqrt{45}< \sqrt{49}< 7\)
Do đó : \(\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)
c) Ta thấy : \(\sqrt{19}>\sqrt{16}=4\)
\(\Rightarrow2\sqrt{19}>2.4=8\)
\(\Rightarrow-2\sqrt{19}< -8\)
\(\Rightarrow23-2\sqrt{19}< 23-8=15\)
\(\Rightarrow\frac{23-2\sqrt{19}}{3}< 5\). Mặt khác : \(\sqrt{27}>\sqrt{25}=5\)
Nên : \(\frac{23-2\sqrt{19}}{3}< \sqrt{27}\)
d) Vì : \(18>12>0\Rightarrow\sqrt{18}>\sqrt{12}>0\)
\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}>0\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Xét hiệu \(\left(\sqrt{2}+\sqrt{6}\right)-\left(\sqrt{3}+2\right)\)
\(=\sqrt{6}-\sqrt{3}+\sqrt{2}-2\)
\(=\sqrt{2}.\sqrt{3}-\sqrt{3}+\sqrt{2}-\sqrt{2}.\sqrt{2}\)
\(=\sqrt{3}.\left(\sqrt{2}-1\right)-\sqrt{2}.\left(\sqrt{2}-1\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{2}-1\right)>\left(\sqrt{2}-\sqrt{2}\right)\left(\sqrt{1}-1\right)=0\)
Hay \(\sqrt{2}+\sqrt{6}>\sqrt{3}+2\)
Ta có :
\(\sqrt{2}+6\)
\(=\sqrt{2}+2+4\)
\(=\sqrt{2}+2+\sqrt{2}\)
\(=\left(\sqrt{2}\right)^2+2\)(1)
Và \(\sqrt{3}+2\)(2)
Từ (1) và (2)
\(\Rightarrow\sqrt{3}+2< \left(\sqrt{2}\right)^2+2\)
\(\Rightarrow\sqrt{3}+2< \sqrt{2}+6\)
Vậy .............