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

help me ai nhanh nhất mik tích cho

a) Ta có: \(\left(\dfrac{3}{4}\right)^{2021}>\left(\dfrac{3}{4}\right)^1=\dfrac{3}{4}\)

\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{2021}+1>\dfrac{3}{4}+1\)

3:

a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

2:

a: 2căn 7=căn 28

3căn 2=căn 18

mà 28>18

nên 2*căn 7>3*căn 2

b: 5=2+3

mà 3>căn 2

nên 2+3>2+căn 2

=>5>2+căn 2

1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)

\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)

\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)

2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)

\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)

Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)

b) \(5=2+3=2+\sqrt{9}\)

Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)

3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

??

a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)

Áp dụng bổ đề trên ta có:

\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)

b)

TH1: x chẵn  mà x là số nguyên tố => x=2

=> y^2 = 117+4=121 => y=11 (thỏa mãn)

TH2:  x lẻ => x^2 lẻ  . Mà 117 lẻ

=> x^2+117 chẵn => y^2 chẵn => y chẵn mà y là số nguyên tố

=> y=2 

=>x^2+117= 4=> x^2 = -113 (vô lý)

Vậy x=2;y=11

a) Rút gọn được \(\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

c) \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\Rightarrow H^2=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}\)

\(\Rightarrow H^2-H=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}-\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}=\dfrac{xy-\sqrt{xy}\left(x-\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}\)

\(=\dfrac{2xy-x\sqrt{xy}-y\sqrt{xy}}{\left(x-\sqrt{xy}+y\right)^2}=\dfrac{-\sqrt{xy}\left(x-2\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\)

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(x-\sqrt{xy}+y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow H^2-H=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\le0\Rightarrow H^2\le H\)

Mà \(H\ge0\left(cmt\right)\Rightarrow H\le\sqrt{H}\)

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)

\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)

\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)

\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)