\(\left(\sqrt{26}+3\right)^2=35+6\sqrt{26}\)

\(\left(\sqrt{63}\right)^2=63=35+28\)

mà \(6\sqrt{26}>28\)

nên \(\sqrt{26}+3>\sqrt{63}\)

\(\sqrt{3}>\sqrt{2}\Rightarrow2\sqrt{2}\cdot\sqrt{3}>4\Rightarrow3+2\sqrt{3}\sqrt{2}+2>9\)

\(\Rightarrow\left(\sqrt{3}+\sqrt{2}\right)^2>3^2\Rightarrow\sqrt{3}+\sqrt{2}>3\Rightarrow\sqrt{3}>3-\sqrt{2}\)

\(\Rightarrow7+\sqrt{3}>10-\sqrt{2}\)

Giả sử \(8>\sqrt{15}+\sqrt{17}\)

\(\Leftrightarrow64>32+2\sqrt{15×17}\)

\(\Leftrightarrow16>\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\left(dung\right)\)

Vậy \(8>\sqrt{15}+\sqrt{17}\)

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căn 15 < căn 16=4

căn 8 < căn 9 bằng 3 

mà 4=3=7 suy ra 7>căn 15 cộng căn 8

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b: \(\sqrt{\dfrac{3}{2}}>\sqrt{\dfrac{2}{2}}=1\)

a: \(\left(2\sqrt{5}-3\sqrt{2}\right)^2=38-12\sqrt{10}=1+37-12\sqrt{10}\)

\(1^2=1\)

mà \(37-12\sqrt{10}< 0\)

nên \(2\sqrt{5}-3\sqrt{2}< 1\)

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.

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

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

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

Đầu tiên ta bình phương tất cả:

\(\sqrt{3^2}=3\)

\(5^2=25\)

\(\sqrt{8^2}=8\)

Sau khi bình phương ta có:

   3 ... 25 - 8

   3 < 17

=> \(\sqrt{3}< 5-\sqrt{8}\)