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 => \(\sqrt{256}>\sqrt{255}\)

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

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

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

8 lớn hơn

\(\sqrt{15}+\sqrt{17}\approx7,9961\)

=> \(8>\sqrt{15}+\sqrt{17}\)

hok tốt 

==.==

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 \(4\sqrt{5}>5\sqrt{3}\)

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:

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

Bài này dễ lắm

Câu 1

\(-\sqrt{5}\) lớn hơn \(-2\) . Vì 

\(-\sqrt{5}=-2,2236067977\) 

\(-2=-2\) 

Câu 2

\(\sqrt{2}+\sqrt{3}\) bé hơn \(\sqrt{10}\) . Vì

\(\sqrt{2}+\sqrt{3}=3,146264\)

\(\sqrt{10}=3,16227766\) 

Câu 3

\(8\) lớn hơn \(\sqrt{15}+\sqrt{17}\) 

\(8=8\)

\(\sqrt{15}+\sqrt{17}=7,996088972\)

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

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

a) 1,2+3.1,3=5,1

b) 0,2+2.0,5=1,2

a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)

Bài 1: 

a) \(\sqrt{1.44}+3\sqrt{1.69}=1.2+3\cdot1.3=1.2+3.9=5.1\)

b) \(\sqrt{0.04}+2\cdot\sqrt{0.25}=0.2+2\cdot0.5=1.2\)

a) Ta có: \(2=\sqrt{4}\)

Vì \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\Rightarrow2>\sqrt{3}\Rightarrow1>\sqrt{3}-1\)

b) \(\left\{{}\begin{matrix}2\sqrt{31}=\sqrt{4.31}=\sqrt{124}\\10=\sqrt{100}\end{matrix}\right.\)

Vì \(124>100\Rightarrow\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)

c) Vì \(15< 16\Rightarrow\sqrt{15}< \sqrt{16}\Rightarrow\sqrt{15}-1< \sqrt{16}-1\)

\(\Rightarrow\sqrt{15}-1< 4-1\Rightarrow\sqrt{15}-1< 3\)

Lại có: \(10>9\Rightarrow\sqrt{10}>\sqrt{9}\Rightarrow\sqrt{10}>3\)

\(\Rightarrow\sqrt{10}>\sqrt{15}-1\)

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