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:

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

a: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)

\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)

mà 49>43 và 528>406

nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)

=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

a) Ta có: \(4+\sqrt{33}=\sqrt{16}+\sqrt{33}\)

Vì \(\sqrt{16}>\sqrt{14};\sqrt{33}>\sqrt{29}\)

\(\Rightarrow4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

b) Ta có: \(\sqrt{23}+\sqrt{15}< \sqrt{25}+\sqrt{16}=5+4=9=\sqrt{81}\)

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

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

\(=3+\sqrt{5}+3-\sqrt{5}=6\)

c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

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

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

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

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

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)

\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)

a, Ta có

\(7^2=49\)

\(\sqrt{42}^2=42\)

\(\Rightarrow\sqrt{42}< 7\)

b, Ta có

\(\sqrt{12}+\sqrt{35}\Leftrightarrow\sqrt{12^2}+\sqrt{35^2}=12+35=47\)

\(6+\sqrt{21}\Leftrightarrow6^2+\sqrt{21^2}=36+21=57\)

\(\Rightarrow\sqrt{12}+\sqrt{35}< 6+\sqrt{21}\)

\(c,\)Ta có

\(4+\sqrt{33}\Leftrightarrow16+\sqrt{33^2}=16+33=49\)

\(\sqrt{29}+\sqrt{14}\Leftrightarrow\sqrt{29^2}+\sqrt{14^2}=29+14=43\)

\(\sqrt{29}+\sqrt{14}< 4+\sqrt{33}\)

Câu d làm nốt nhé lười lắm. Không biết có sai k nếu sai thì chỉ cho mik vs nhé mn

a, Ta có: \(\sqrt{49}>\sqrt{42}\Leftrightarrow7>\sqrt{42}\)

b, Ta có: \(\sqrt{12}+\sqrt{35}< \sqrt{21}+\sqrt{36}=\sqrt{21}+6\)

c, Ta có: \(4+\sqrt{33}=\sqrt{16}+\sqrt{33}>\sqrt{14}+\sqrt{29}\)

d, Ta có: \(\sqrt{48+\sqrt{149}}< \sqrt{48+\sqrt{169}}=\sqrt{48+13}=\sqrt{61}< \sqrt{324}=18\)

Mk gợi ý vậy thôi bn tự trình bày nhé
STD well

a) Ta có: \(\sqrt{42}=6,48...\)

Vì 7 > 6,48... => \(7>\sqrt{42}\)

b) Ta có: \(\sqrt{12}+\sqrt{35}=9,38...\)

               \(6+\sqrt{21}=10,58...\)

Vì \(9,38...< 10,58...\rightarrow\sqrt{12}+\sqrt{35}< 6+\sqrt{21}\)

c) Ta có:\(4+\sqrt{33}=9,74...\)

            \(\sqrt{29}+\sqrt{14}=9,12...\)

Vì \(9,74...>9,12...\rightarrow4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

d) Ta có:\(\sqrt{48+\sqrt{149}}=7,75...\)

Vì \(7,75...< 18\rightarrow\sqrt{48+\sqrt{149}}< 18\)