ta thấy \(\sqrt{65}>\sqrt{64}\Leftrightarrow\sqrt{65}-1>\sqrt{64}-1\)

mà ta có \(\sqrt{64}-1=8-1=4+3=\sqrt{16}+\sqrt{9}\)

lại có \(\sqrt{16}>\sqrt{15};\sqrt{9}>\sqrt{8}\Leftrightarrow\sqrt{16}+\sqrt{9}>\sqrt{15}+\sqrt{8}\)

Vậy \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)

Vậy \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

\(\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{25}=3+5=8=\sqrt{64}=\sqrt{65-1}\)

\(\sqrt{65-1}=\sqrt{64}=8\)

\(\sqrt{7}<\sqrt{9};\sqrt{15}<\sqrt{16}\rightarrow\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{16}=3+4=7<8\)

Do đó phải điền dấu < 

a)

Ta có

\(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)

b) Ta có

\(\sqrt{17}+\sqrt{5}+9>\sqrt{16}+\sqrt{4}+9=4+2+9=15\)

\(\Rightarrow\sqrt{17}+\sqrt{5}+9>15\)

Mặt khác

\(\sqrt{115}< \sqrt{225}=15\)

Mà \(\sqrt{17}+\sqrt{5}+9>15\)

\(\Rightarrow\sqrt{115}< \sqrt{17}+\sqrt{5}+9\)

ta có \(\sqrt{7}< \sqrt{9}\)

và \(\sqrt{15}< \sqrt{16}\)

=> \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

mà \(\sqrt{9}+\sqrt{16}=3+4=7\)

=> \(\sqrt{7}+\sqrt{15}< 7\)

a) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\sqrt{65}-1>\sqrt{64}-1=8-1=7\)

\(\Rightarrow\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

b) \(\frac{13-2\sqrt{3}}{6}>\frac{13-2\sqrt{4}}{6}=1,5\)

mà 1,5² = 2,25 ; \(\sqrt{2}^2=2\)

\(\Rightarrow1,5>\sqrt{2}\)hay \(\frac{13-2\sqrt{3}}{6}>\sqrt{2}\)

\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

Lời giải:

a.

$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$

$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.

$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$

$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$

$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$

Ta có

\(\sqrt{123-22\sqrt{2}}=11-\sqrt{2}\)

\(\sqrt[3]{77\sqrt{2}-115}=\sqrt{2}-5\)

\(\Rightarrow\sqrt{123-22\sqrt{2}}+\sqrt[3]{77\sqrt{2}-115}=11-\sqrt{2}+\sqrt{2}-5=6\)

\(\sqrt{3}+\sqrt{15}< \sqrt{5}+\sqrt{16}=\sqrt{5}+4\)