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 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà

2/ 

a) Ta có:

\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)

\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)

Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)

b) Ta có:

\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)

\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)

Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)

3/

a)ĐKXĐ: \(x\ne1;x\ge0\)

b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)

\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)

\(A=1^2-\left(\sqrt{x}\right)^2\)

\(A=1-x\)

1/ \(\sqrt[3]{54}-\sqrt[3]{16}\)

\(=\sqrt[3]{3^3\cdot2}-\sqrt[3]{2^3\cdot2}\)

\(=3\sqrt[2]{3}-2\sqrt[3]{2}\)

\(=\left(3-2\right)\sqrt[3]{2}\)

\(=\sqrt[3]{2}\)

a) 7 và \(\sqrt{37}+1\)

=7 và 7,08

=>......

b) \(\sqrt{17}-\sqrt{50}-1\)và \(\sqrt{99}\)

=-3,95 và 9,95

=>.....

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:

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

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

2) \(-x^2+4x-2\)

\(=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)\)

\(=-\left(x-2\right)^2+2\)

Ta có: \(-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)

Dấu "=" xảy ra:

\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)

Vậy: GTLN của bt là 2 tại x=2

b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))

Mà: \(\sqrt{2x^2-3}\ge0\forall x\)

Dấu "=" xảy ra:

\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)

Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)

...

1:

b: \(4\sqrt{5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{75}\)

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

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

c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)

=>\(3-2\sqrt{5}< 1-\sqrt{5}\)

d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)

\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)

=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)

=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)

e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)

\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)

=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)

=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)