Ta có:

\(\left(\sqrt{65}+\sqrt{63}\right)^2=128+2\sqrt{65.63}< 128+2\sqrt{64.64}=\left(\sqrt{64}+\sqrt{64}\right)^2\Rightarrow\sqrt{65}+\sqrt{63}< \sqrt{64}+\sqrt{64}\)

\(\Rightarrow a=\sqrt{65}-\sqrt{63}=\frac{2}{\sqrt{65}+\sqrt{63}}>\frac{2}{\sqrt{64}+\sqrt{64}}=\frac{2}{2\sqrt{64}}=\frac{1}{8}\)

Ta có:

\(a=\frac{2}{\sqrt{65}+\sqrt{63}}< \frac{2}{\sqrt{64}+\sqrt{49}}=\frac{2}{15}\)

Vậy \(\frac{1}{8}< a< \frac{2}{15}\)

đề sai

\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}= \sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{5+2\sqrt{3}.\sqrt{5}+3}\\ =\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\)

+) \(\left(\sqrt{4}-\sqrt{3}\right)^2=4-2\sqrt{4\cdot3}+3=7-2\sqrt{7}=\sqrt{49}-\sqrt{48}\)

+) \(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}\)

\(=4\sqrt{2}-6\sqrt{6}+9-4\sqrt{2}+6\sqrt{6}\)

\(=9\)

+) Sửa : \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

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

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

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

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

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 \(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)

Sửa đề: Chứng minh \(\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)=\left(\sqrt{3}-1\right)^2\)

Ta có: \(VT=\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)\)

\(=\left(\sqrt{4+2\cdot2\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\right)-\left(\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\right)\)

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

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

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

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

\(=4-2\sqrt{3}\)

\(=3-2\cdot\sqrt{3}\cdot1+1\)

\(=\left(\sqrt{3}-1\right)^2=VP\)(đpcm)

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.

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

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

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

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

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

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

Bạn Nguyễn Hà Vy là đúng rồi, chỉ hơi nhầm (viết thiếu) khi viết căn bậc hai của 9 thôi.

Trình bày lại bài làm của bạn Hà Vy như sau:

\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7=8-1< \sqrt{64}-1< \sqrt{65}-1\)

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