\(1.-3< -5+\sqrt{5}\)

\(2.-4>-2\sqrt{5}\)

\(3.-3\sqrt{5}< -6\)

2) \(4=\sqrt{16}\)

\(2\sqrt{5}=\sqrt{20}\)

mà 16<20

nên \(-4>-2\sqrt{5}\)

3) \(3\sqrt{5}=\sqrt{45}\)

\(6=\sqrt{36}\)

mà 45>36

nên \(-3\sqrt{5}< -6\)

1)Ta có \(-3=-\sqrt{9}>-5+\sqrt{5}\)

2)Ta có \(-2\sqrt{5}=(-\sqrt{20})<-4=(-\sqrt{16})\)

3)Ta có \(-3\sqrt{5}=(-\sqrt{45})<-6=-\sqrt{36}\)

\(\sqrt{3}+\sqrt{15}< \sqrt{5}+\sqrt{16}=\sqrt{5}+4\)

Đặt: 

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

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(\left|1+\sqrt{5}\right|+\left|\sqrt{5}-1\right|\right)\)

\(A=\dfrac{1}{\sqrt{2}}\left(1+\sqrt{5}+\sqrt{5}-1\right)\)

\(A=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

Ta có: \(A^2=\left(\sqrt{10}\right)^2=10\)  

\(B=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

Mà: \(4\sqrt{5}>1\)

Nên: \(A^2< B^2\)

\(\Rightarrow A< B\)

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1+\sqrt{5}-1\right)=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

=>A^2=(căn 10)^2=10=9+1

Đặt B=2+căn 5

=>B^2=(2+căn 5)^2=9+4căn 5

1<4căn 5

=>9+1<9+4căn 5

=>A^2<B^2

=>A<B

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

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

\(=6+2\sqrt{9-5}=6+2.2=10\)

\(B=2+\sqrt{5}\Rightarrow B^2=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

\(>9+1=10=A^2\)

\(\Rightarrow B^2>A^2\Rightarrow B>A\)

Vậy, B>A

Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

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:

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

\(\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}\)\(\approx8,382332347\)\(>6\)

Ta có:\(\sqrt{2}>1;\sqrt{3}>1;\sqrt{4}>1;\sqrt{5}>2\)

=>\(1+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}>1+1+1+1+2=6\)

=>đpcm

Ta có: \(\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}\) 

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

\(\Rightarrow1+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}>1+1+1+1+2=6\)

\(\Rightarrowđpcm\)