\(\sqrt{99}<\sqrt{100}=10\)

\(\sqrt{50}+\sqrt{10}\)\(>\sqrt{49}+\sqrt{9}=7+3=10\)

Vậy \(\sqrt{50}+\sqrt{10}>\sqrt{99}\)

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 ( * ) ; ( ** ) 

\( \implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{64}}>14\left(đpcm\right)\) 

\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)

\(=6+3-7\)

\(=2\)

\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)

\(=\sqrt{2}\cdot\left(5\sqrt{2}-3\sqrt{2}\right)\)

\(=\sqrt{2}\cdot2\sqrt{2}\)

\(=4\)

a) \(T=\sqrt{36}+\sqrt{9}-\sqrt{49}\)

    \(=6+3-7\)

      \(=2\)

b)  \(B=\sqrt{2\left(\sqrt{50}-3\sqrt{2}\right)}\)

         \(=\sqrt{10\sqrt{2}-6\sqrt{2}}\)

           \(=\sqrt{\left(10-6\right)\sqrt{2}}\)

           \(=\sqrt{4\sqrt{2}}\)

            \(\approx2,39\)

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 

\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-12\sqrt{5}+4}-\sqrt{45+12\sqrt{5}+4}\)

\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)

\(=|3\sqrt{5}-2|-|3\sqrt{5}+2|\)

\(=3\sqrt{5}-3\sqrt{5}-4=-4\)

4

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

ban lam co dung ko vay

nếu đề của bạn nêu là âm căn 49 thì mình làm đúng nha