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

Giúp mk vs, mk cần gấp!!!

a: \(324=48+276=48+\sqrt{76176}>48+\sqrt{120}\)

nên \(\sqrt{48+\sqrt{120}}< 18\)

b: \(\left(\sqrt{23}+\sqrt{15}\right)^2=38+2\cdot\sqrt{345}\)

\(\left(\sqrt{91}\right)^2=91=38+53=38+\sqrt{2809}\)

mà \(2\sqrt{345}< \sqrt{2809}\)

nên \(\sqrt{23}+\sqrt{15}< \sqrt{91}\)

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

\(7^2=49\)

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

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

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

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 \(4\sqrt{5}>5\sqrt{3}\)

Giải:

Vì \(4=\sqrt{16}\Rightarrow4+\sqrt{33}=\sqrt{16}+\sqrt{33}.\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{16}>\sqrt{14}_{\left(1\right).}\\\sqrt{33}>\sqrt{29}_{\left(2\right).}\end{matrix}\right.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) suy ra: \(\sqrt{16}+\sqrt{33}>\sqrt{29}+\sqrt{14}.\)

Hay: \(4+\sqrt{33}>\sqrt{29}+\sqrt{14}.\)

Vậy.....

~ Học tốt!!! ~

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:

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