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 \(4\sqrt{5}>5\sqrt{3}\)

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 \(3\sqrt{12}>2\sqrt{16}\)

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:

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

1) `-3\sqrt13=-3\sqrt13`

`-9=-3\sqrt9`

`\sqrt13>\sqrt9`

`=> -3\sqrt13 < -3\sqrt9`

`=> -3\sqrt13 < 9`.

2) `\sqrt15 < \sqrt16`

`<=> \sqrt15-1 < \sqrt16-1`

`<=> \sqrt15-1 < 3 < \sqrt10`

`=> \sqrt15-1 <\sqrt10`

3) `5=4+1=\sqrt16+1`

`\sqrt8+1=\sqrt8+1`

`=> 5>\sqrt8+1`

1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)

3) Ta có: \(5^2=25=9+16\)

\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)

mà \(16>4\sqrt{2}\)

nên \(5>2\sqrt{2}+1\)

1) $-3\sqrt{13}=-\sqrt{117}<-\sqrt{81}=-9$

3) Ta có: ${}^{}=25=9+16$

${}^{}=9+4\sqrt{2}$

mà $16>4\sqrt{2}$

nên $5>2\sqrt{2}+1$

a    \(\left(\sqrt{5\sqrt{7}}\right)^4=\left(\left(\sqrt{5\sqrt{7}}\right)^2\right)^2=\left(5\sqrt{7}\right)^2=25\cdot7=175\)

\(=\left(\sqrt{7\sqrt{5}}\right)^4=\left(\left(\sqrt{7\sqrt{5}}\right)^2\right)^2=\left(7\sqrt{5}\right)^2=49\cdot5=240\)

vì 175<240\(\Rightarrow\left(\sqrt{5\sqrt{7}}\right)^4< \left(\sqrt{7\sqrt{5}}\right)^4\Rightarrow\sqrt{5\sqrt{7}}< \sqrt{7\sqrt{5}}\)

b     \(6=\sqrt{36}\)

\(\sqrt{31}< \sqrt{36};\sqrt{19}>\sqrt{17}\Rightarrow\sqrt{31}-\sqrt{19}< \sqrt{36}-\sqrt{17}=6-\sqrt{17}\)

c      \(\left(\sqrt{10}+\sqrt{17}\right)^2=10+2\sqrt{10\cdot17}+17=27+2\sqrt{170}\)

\(\left(\sqrt{61}\right)^2=61=27+34=27+2\cdot17=27+2\sqrt{289}\)

vì \(2\sqrt{170}< 2\sqrt{289}\Rightarrow27+2\sqrt{170}< 27+2\sqrt{289}\Rightarrow\left(\sqrt{10}+\sqrt{17}\right)^2< \left(\sqrt{61}\right)^2\)

\(\Rightarrow\sqrt{10}+\sqrt{17}< \sqrt{61}\)

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

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