Ta có \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2021}\right)\left(1-\dfrac{1}{2022}\right)\)

\(B=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}.\dfrac{2021}{2022}\)

\(B=\dfrac{1}{2022}\)

giúp mik với ạ

B = (1 - \(\dfrac{1}{2}\))(1 - \(\dfrac{1}{3}\))(1 - \(\dfrac{1}{4}\))...(1-\(\dfrac{1}{2021}\))(1 - \(\dfrac{1}{2022}\))

B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)...\(\times\)\(\dfrac{2021-1}{2021}\)\(\times\)\(\dfrac{2022-1}{2022}\)

B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)...\(\times\)\(\dfrac{2020}{2021}\)\(\times\)\(\dfrac{2021}{2022}\)

B = \(\dfrac{2\times3\times4\times...\times2021}{2\times3\times4\times...\times2021}\) \(\times\) \(\dfrac{1}{2022}\)

B = \(\dfrac{1}{2022}\)

=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023

=0+0+...+0-1-2023

=-2024

Bằng 2024

P(k)=1/k+1

=>P(2023)=1/2023+1=1/2024

3: 

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

Cuối cùng trong năm 2021 thôi nhé.

Đặt biểu thức trên là A

TC

√1 + 1/1^2 + 1/2^2 = 1 + 1 - 1/2

Tương tự

√1 + 1/2^2 + 1/3^2 = 1 + 1/2 -  1/3

√1 + 1/2021^2 + 2022^2 = 1 + 1/2021 -  1/2022

=> A = (1 + 1 + 1/3 +...+ 1/2021) - (1/2 + 1/3 +....+ 1/2022)

=> A = 1 + 1 - 1/2022 = 4043/2022

đúng không bạn

Dựa theo dạng này nha em

Với \(a+b+c=0\left(abc\ne0\right)\) ta có :

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Áp dụng cho từng thừa số A ( anh gọi biểu thức này là A ) , ta có :

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{\left(1+\frac{1}{2}+\frac{1}{-3}\right)-2\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)}\)

\(=\sqrt{\left(1+\frac{1}{2}+\frac{1}{-3}\right)-2.\frac{3-2-1}{6}}=1+\frac{1}{2}+\frac{1}{3}\)

Tương tự : \(\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=1+\frac{1}{3}+\frac{1}{4}\)

\(\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=1+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)\)

\(\Rightarrow1+\frac{1}{2}-\frac{1}{100}=\frac{149}{100}\)

-Mình làm tắt được không bạn :/?

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}\)

\(\Rightarrow\dfrac{bc+ca+ab}{abc}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\left(bc+ca+ab\right)\left(a+b+c\right)=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)

\(\Rightarrow c=2022\) hay \(a=2022\) hay \(b=2022\)

-Nếu \(a=-b\)\(\Rightarrow B=\dfrac{1}{a^{2021}}+\dfrac{1}{b^{2021}}+\dfrac{1}{c^{2021}}=\dfrac{1}{a^{2021}}-\dfrac{1}{a^{2021}}+\dfrac{1}{2022^{2021}}=\dfrac{1}{2022^{2021}}\)

-Tương tự các trường hợp còn lại.

ok

bye

\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)

\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)

\(=0\)

Ta có :                  

                ^{82 - 576 : 32}

^{= 64 - 576 : 9}

^{= 64 - 64}

^{=  0}

 (1¹ + 2² + 3³ + 4⁴ +...+ 2022²⁰²²) . 0

^{= 0}