chứng minh căn 50 + căn 49 >14
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{99}<\sqrt{100}=10\)
\(\sqrt{50}+\sqrt{10}\)\(>\sqrt{49}+\sqrt{9}=7+3=10\)
Vậy \(\sqrt{50}+\sqrt{10}>\sqrt{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận thấy với mọi k \(\in\) N* ta có :
\(\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)=\left(\sqrt{k+1}\right)^2-\left(\sqrt{k}\right)^2=k+1-k=1\)
\( \implies\)\(\frac{\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}\)
\( \implies\) \(\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k}\)
Thật vậy : \(\frac{1}{\sqrt{k}}=\frac{2}{2.\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2.\left(\sqrt{k+1}-\sqrt{k}\right)\)
Thay k = 1 ; 2 ; 3 ; ....; 64 ta được :
\(\frac{1}{\sqrt{1}}>2.\left(\sqrt{1+1}-\sqrt{1}\right)=2.\left(\sqrt{2}-\sqrt{1}\right)=2.\sqrt{2}-2.\sqrt{1}\)
\(\frac{1}{\sqrt{2}}>2.\left(\sqrt{2+1}-\sqrt{2}\right)=2.\left(\sqrt{3}-\sqrt{2}\right)=2.\sqrt{3}-2.\sqrt{2}\)
\(\frac{1}{\sqrt{3}}>2.\left(\sqrt{3+1}-\sqrt{3}\right)=2.\left(\sqrt{4}-\sqrt{3}\right)=2.\sqrt{4}-2.\sqrt{3}\)
. . . . . . . . . . . . . . . . . . . . . .
\(\frac{1}{\sqrt{64}}>2.\left(\sqrt{64+1}-\sqrt{64}\right)=2.\left(\sqrt{65}-\sqrt{64}\right)=2.\sqrt{65}-2.\sqrt{64}\)
Cộng vế với vế ta được :
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+....+2.\sqrt{65}-2.\sqrt{64}\)
\( \implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{65}-2.\sqrt{1}=2.\left(\sqrt{65}-\sqrt{1}\right)\) ( * )
Ta thấy : \(\sqrt{65}>\sqrt{64}\)
\( \implies\) \(\sqrt{65}-\sqrt{1}>\sqrt{64}-\sqrt{1}\)
\( \implies\) \(\sqrt{65}-\sqrt{1}>7\)
\( \implies\) \(2.\left(\sqrt{65}-\sqrt{1}\right)>2.7\)
\( \implies\) \(2.\left(\sqrt{65}-\sqrt{1}\right)>14\) ( ** )
Từ ( * ) ; ( ** )
\( \implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{64}}>14\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)
\(=6+3-7\)
\(=2\)
\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\left(5\sqrt{2}-3\sqrt{2}\right)\)
\(=\sqrt{2}\cdot2\sqrt{2}\)
\(=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-12\sqrt{5}+4}-\sqrt{45+12\sqrt{5}+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2\)
\(=-4\)
Học tốt
\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-12\sqrt{5}+4}-\sqrt{45+12\sqrt{5}+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=|3\sqrt{5}-2|-|3\sqrt{5}+2|\)
\(=3\sqrt{5}-3\sqrt{5}-4=-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
âm căn 49=-7, căn 25 bằng 5
=>-7<x<5
mà x chia hết cho 2 => x chẵn
=> x thuộc tập hợp: -6,-4,-2,0,2,4
CM:căn 50 + căn 49 >14
căn 50+7>14
căn 50>7
mà căn 50> căn 49=7
=> căn 50 + căn 49>14
k cho mình nha