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 2 : 

a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)

b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)

c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)

Bài 1 :

\(c,\sqrt{15}.\sqrt{17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}.\)

\(16=\sqrt{16^2}\)\(\Leftrightarrow16>\sqrt{15}.\sqrt{17}\)

Câu d coi lại đề giùm :> 

Bài 2 : 

\(a,\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{7}}{2\sqrt{3}+2\sqrt{7}}\)

\(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)

\(b,\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(\sqrt{2}+1\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

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/ \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2.3}=5+2\sqrt{6}=5+\sqrt{24}\)

\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)

Vì \(\sqrt{24}< \sqrt{25}\)

=>\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)

b/\(\left(\sqrt{3}+2\right)^2=3+4+4\sqrt{3}=7+4\sqrt{3}\)

\(\left(\sqrt{2}+\sqrt{16}\right)^2=2+16+2\sqrt{2.16}=18+4\sqrt{8}\)

=> \(\sqrt{3}+2< \sqrt{2}+\sqrt{16}\)

c/ \(16=\sqrt{16^2}\)

\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)

=> \(16>\sqrt{15}.\sqrt{17}\)

d/\(8^2=64=32+32=32+2\sqrt{256}\)

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

=> \(8>\sqrt{15}+\sqrt{17}\)

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