giả sử \(8>\sqrt{15}+\sqrt{17}\)

\(\Leftrightarrow64>32+2\sqrt{15.17}\)

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

Vậy \(8>\sqrt{15}+\sqrt{17}\)

giải thích thêm cho bạn dễ hiểu:

Ta có: \(\left(\sqrt{15}+\sqrt{17}\right)=15+17+2\sqrt{15.17}\)

                                              \(32+\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)

Ta có: \(8=\sqrt{16}+\sqrt{16}\)

\(\left(\sqrt{16}+\sqrt{16}\right)^2=16+16+2.\sqrt{16.16}=32+2.\sqrt{256}\)

\(\left(\sqrt{15}+\sqrt{17}\right)^2=15+17+2.\sqrt{\left(15.17\right)}=32+2.\sqrt{255}\)

Vì 255 > 256 => \(\sqrt{256}>\sqrt{255}\)

\(\Rightarrow32+2.\sqrt{256}>32+2.\sqrt{255}\)

\(\Rightarrow\sqrt{16}+\sqrt{16}>\sqrt{15}+\sqrt{17}\)

\(\Rightarrow8>\sqrt{15}+\sqrt{17}\)

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

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

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}\)  

\(16=\sqrt{16^2}\)

Ta có:  \(15.17=\left(16-1\right)\left(16+1\right)=16^2-1< 16^2\)

\(\Rightarrow\)\(\sqrt{15.17}< \sqrt{16^2}\)

\(\Rightarrow\) \(\sqrt{15}.\sqrt{17}< 16\)

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

8 lớn hơn \(\sqrt{15}\)+\(\sqrt{17}\)

vì \(\sqrt{15}\)+\(\sqrt{17}\)=7,997,,

16>căn 15 nhân căn 17, do can 5 nhan can 17 =15,968........<16

chúc bn học tốt!!!!!!!

Ta có \(256>255\Leftrightarrow256>15.17\)

                                 \(\Leftrightarrow\sqrt{256}>\sqrt{15.17}\)

                                 \(\Leftrightarrow16>\sqrt{17}.\sqrt{15}\)

Giả sử \(8< \sqrt{15}+\sqrt{17}\)

\(\Leftrightarrow64< 15+2\sqrt{15.17}+17\)(Bình phương hai vế)

\(\Leftrightarrow32< 2\sqrt{15.17}\)

\(\Leftrightarrow16< \sqrt{15.17}\)

\(\Leftrightarrow16< \sqrt{\left(16-1\right)\left(16+1\right)}\)

\(\Leftrightarrow\sqrt{16^2}< \sqrt{16^2-1}\)

\(\Leftrightarrow16^2< 16^2-1\)(vô lí)

Chứng minh tương tự điều giả sử \(8=\sqrt{15}+\sqrt{17}\)

Vậy \(8>\sqrt{15}+\sqrt{17}\)

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bn coppy link này nhé, có bài mak bn đang cần đấy

Ta có:\(8=4+4=\sqrt{16}+\sqrt{16}\)

\(\Rightarrow\left(\sqrt{16}+\sqrt{16}\right)^2=16+16+2\cdot\sqrt{16^2}\)

Mà \(\sqrt{16^2}=\sqrt{15\cdot16+16}>\sqrt{15\cdot16+15}=\sqrt{15\cdot17}\)

Nên suy ra:

\(16+16+2\cdot\sqrt{16^2}=32+2\cdot\sqrt{16^2}>32+2\cdot\sqrt{15\cdot17}=15+17+2\cdot\sqrt{15\cdot17}\)

\(\Leftrightarrow8>\sqrt{15}+\sqrt{17}\)