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)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11

\(12< 17\Rightarrow\sqrt{12}< \sqrt{17}\)

tất nhiên là nhỏ hơn rồi

Vì 12<17 nên \(\sqrt{12}< \sqrt{17}\)

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: \(1< \sqrt{2}\)

nên \(2< \sqrt{2}+1\)

b: \(2\sqrt{31}=\sqrt{124}\)

\(10=\sqrt{100}\)

mà 124>100

nên \(2\sqrt{31}>10\)

c: \(-3\sqrt{11}=-\sqrt{99}\)

\(-\sqrt{12}=-\sqrt{12}\)

mà 99>12

nên \(-3\sqrt{11}< -\sqrt{12}\)

\(A=\left(\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{5}+\sqrt{6}+\sqrt{7}+\sqrt{8}+\sqrt{9}\right)+\left(\sqrt{10}+\sqrt{11}+\sqrt{12}\right)\)

Ta có: 

\(\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}>1+\sqrt{1}+\sqrt{1}+\sqrt{1}+2=5\)

\(\sqrt{5}+\sqrt{6}+\sqrt{7}+\sqrt{8}+\sqrt{9}>\sqrt{5}+\sqrt{5}+\sqrt{5}+\sqrt{5}+\sqrt{5}=5\sqrt{5}\)

\(\sqrt{10}+\sqrt{11}+\sqrt{12}>\sqrt{9}+\sqrt{9}+\sqrt{9}=9\)

=> \(A>5+5\sqrt{5}+9=14+5\sqrt{5}>12+5\sqrt{5}\)

Vậy...

a) Ta có:

\(2=1+1=1+\sqrt{1}\)

Mà: \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\)

\(\Rightarrow1+\sqrt{1}< \sqrt{2}+1\)

\(\Rightarrow2< \sqrt{2}+1\)

b) Ta có:

\(1=2-1=\sqrt{4}-1\)

Mà: \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\)

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

\(\Rightarrow1>\sqrt{3}-1\)

c) Ta có:

\(10=2\cdot5=2\sqrt{25}\)

Mà: \(25< 31\Rightarrow\sqrt{25}< \sqrt{31}\)

\(\Rightarrow2\sqrt{25}< 2\sqrt{31}\)

\(\Rightarrow10< 2\sqrt{31}\)

d) Ta có:

\(-12=-3\cdot4=-3\sqrt{16}\)

Mà: \(16>11\Rightarrow\sqrt{16}>\sqrt{11}\)

\(\Rightarrow-3\sqrt{16}< -3\sqrt{11}\)

\(\Rightarrow-12< -3\sqrt{11}\)

a ) 

\(\sqrt{31}+4< \sqrt{36}+4=10\left(1\right)\)

\(6+\sqrt{17}>6+\sqrt{16}=6+4=10\left(2\right)\)

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow\sqrt{31}+4< 10< 6+\sqrt{17}\)

\(\Rightarrow\sqrt{31}+4< \sqrt{17}+6\)

b ) 

\(\sqrt{3}+\sqrt{2}>\sqrt{1}+\sqrt{1}=2\)

c ) 

\(\sqrt{12+13}=\sqrt{25}=5\left(1\right)\)

\(\sqrt{12}+\sqrt{13}>\sqrt{4}+\sqrt{9}=2+3=5\left(2\right)\)

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow\sqrt{12+13}< \sqrt{12}+\sqrt{13}\)

:V

khó vc