\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)

B = 1/6 + 1/12 + 1/20 + ... + 1/90

B = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10

B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10

B = 1/2 - 1/10

B = 5/10 - 1/10

B = 4/10 = 2/5

Ủng hộ mk nha ♡_♡☆_☆

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 = \(\dfrac{1}{2022}\)

\(B=8x^3+12x^2+6x+1\)

\(=8\left(\dfrac{1}{2}\right)^3+12\left(\dfrac{1}{2}\right)^2+6.\dfrac{1}{2}+1\)

\(=8.\dfrac{1}{8}+12.\dfrac{1}{4}+3+1\)

\(=1+3+4\)

\(=8\)

Để tính giá trị của biểu thức B=8x^3+12x^2+6x+1 tại x=1/2, ta thay giá trị này vào biểu thức.

B = 8(1/2)^3 + 12(1/2)^2 + 6(1/2) + 1
= 8(1/8) + 12(1/4) + 6(1/2) + 1
= 1 + 3 + 3 + 1
= 8

Vậy, giá trị của biểu thức B tại x=1/2 là 8.

Thay \(x=\dfrac{1}{2}\) vào biểu thức trên , ta có : 

\(B=\)\(8.\left(\dfrac{1}{2}\right)^3+12.\left(\dfrac{1}{2}\right)^2+6.\dfrac{1}{2}+1\)

\(=8.\dfrac{1}{8}+12.\dfrac{1}{4}+6.\dfrac{1}{2}+1\)

\(=1+3+3+1\)

\(=4+4\)

\(=8\)

Vậy khi \(x=\dfrac{1}{2}\) thì \(B=8\)

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

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

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

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

\(B=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot2020\cdot2021}{2\cdot3\cdot4\cdot\cdot\cdot2021\cdot2022}\)

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

B=(1-1/2).(1-1/3).(1-1/4)...(1-1/2021).(1-1/2022)

B = 1/2 . 2/3 . 3/4 ... 2020/2021 . 2021/2022

B = \(\dfrac{1.2.3.....2020.2021}{2.3.4......2021.2022}\)

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

b) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)

\(=2-\sqrt{3}+\sqrt{3}-1-6\cdot\dfrac{4}{\sqrt{3}}\)

\(=1-8\sqrt{3}\)

5/3

= 1/3 : 1/5 = 1/3 x 5 = 5/3

5\3