A=12/15 + 28/315

            A=8/9 

\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)

\(\frac{10}{11}.y=\frac{4}{3}\)

\(\Rightarrow y=\frac{22}{15}\)

Mình nghĩ là bài 1: 2/141 đổi thành 2/143 mới đúng đề nha.

Bài 1: Tham khảo https://olm.vn/hoi-dap/question/40337.html

ê bạn ơi tìm y sao lại có x

\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)

\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)

\(\frac{2}{3}\cdot y=\frac{10}{3}\)

\(y=\frac{10}{3}:\frac{2}{3}\)

y=5

dấu nhân bạn ạ

\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)

hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)

a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A=2+2^2+2^3+2^4+...+2^{201}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A-A=2^{101}-1\)

\(A=2^{201}-1\)

Ta có 2²⁰¹ > 2²⁰¹ - 1 => B > A => 2²⁰¹ > 1 + 2 + 2² + 2³ +...+ 1¹⁰⁰

b) 2¹⁰⁰ = 2³¹ . 2⁶³ . 2⁶ = 2³¹ . (2⁹)⁷ . (2²)³ = 2³¹ . 512⁷ . 4³ (1)

10³¹ = 2³¹ . 5²⁸ . 5³ = 2³¹ . (5⁴)⁷ . 5³ = 2³¹ . 625⁷ . 5³ (2)

Từ (1) , (2) => 2³¹ . 512⁷ . 4³ < 2³¹ . 625⁷ . 5³ ( vì 512⁷ < 625⁷ và 4³ < 5³ )

=> 2¹⁰⁰ < 10³¹

e) Ta có:

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰

10³⁰ = (10³)¹⁰ = 1000¹⁰
Vì 1024¹⁰ > 1000¹⁰ => 2¹⁰⁰ > 10³⁰

cái này tính cái gì thế

ko hiểu

T nghĩ đề là phép + chứ nhỉ?! phép trừ thì s lm đc?!

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\)

\(=\dfrac{1}{2}+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{201}{202}\)

p/s: Nghĩ vậy còn đề là trừ thì ~~ Chịu ~~

\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)

\(=\dfrac{1}{2}-\dfrac{50}{101}\)

\(=\dfrac{1}{202}.\)