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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)
=10
Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)
\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)
\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)
\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{3\sqrt{2}}=\sqrt{\sqrt{3^2\cdot2}}=\sqrt{\sqrt{18}}\)
\(\sqrt{2\sqrt{3}}=\sqrt{\sqrt{2^2\cdot3}}=\sqrt{\sqrt{12}}\)
từ trên ta suy ra
\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2=\sqrt{4}\)
Vì \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\Rightarrow2>\sqrt{3}\Rightarrow1>\sqrt{3}-1\)
b) \(\left\{{}\begin{matrix}2\sqrt{31}=\sqrt{4.31}=\sqrt{124}\\10=\sqrt{100}\end{matrix}\right.\)
Vì \(124>100\Rightarrow\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)
c) Vì \(15< 16\Rightarrow\sqrt{15}< \sqrt{16}\Rightarrow\sqrt{15}-1< \sqrt{16}-1\)
\(\Rightarrow\sqrt{15}-1< 4-1\Rightarrow\sqrt{15}-1< 3\)
Lại có: \(10>9\Rightarrow\sqrt{10}>\sqrt{9}\Rightarrow\sqrt{10}>3\)
\(\Rightarrow\sqrt{10}>\sqrt{15}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Có: \(2>1>0\)
\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)
b) Có: \(0< \sqrt{3}< 3\)
\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)
c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)
\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)
d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)
\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)
a: 2=1+1<1+căn 2
b: 4=1+3>1+căn 3
c: -2căn 11=-căn 44
-10=-căn 100
mà 44<100
nên -2 căn 11>-10
d: 12=3*4=3*căn 16>3*căn 11
![](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 \(4\sqrt{5}>5\sqrt{3}\)
Giả sử \(\sqrt{7}-\sqrt{2}>1\\ \)
<=> 9 - \(2\sqrt{14}\)> 1
<=> 8 > \(2\sqrt{14}\)
<=> \(64>56\)(đúng)
Vậy \(\sqrt{7}-\sqrt{2}>1\)
Câu còn lại tương tự