2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

Nội qui tham gia "Giúp tôi giải toán"

1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;

2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.

3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.

Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web. 

mong các bn đừng làm như vậy nha

2: \(M=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>3M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}\)

=>3M+M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>4M=\(1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>3A=-1+\(\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>4A=\(-1-\frac{1}{3^{29}}=\frac{-3^{29}-1}{3^{29}}\)

=>\(A=\frac{-3^{29}-1}{4\cdot3^{29}}\)

Ta có: \(4M=1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

\(=1+\frac{-3^{29}-1}{4\cdot3^{29}}-\frac{30}{3^{30}}=1+\frac{-3^{30}-3-120}{4\cdot3^{30}}=1-\frac14-\frac{123}{4\cdot3^{30}}=\frac34-\frac{123}{4\cdot3^{30}}\)

=>4M<3/4

=>M<3/16

a: \(-3\frac12x-0,75-1,25x=\left(-\frac12\right)^2:\frac{-3}{4}+\frac16\)

=>\(-3,5x-1,25x=\frac14\cdot\frac{4}{-3}+\frac16+\frac34\)

=>\(-4,75x=\frac{-1}{3}+\frac16+\frac34=\frac{-4}{12}+\frac{2}{12}+\frac{9}{12}=\frac{-4+11}{12}=\frac{7}{12}\)

=>\(x=\frac{7}{12}:\frac{-19}{4}=\frac{7}{12}\cdot\frac{-4}{19}=\frac{-7}{3\cdot19}=-\frac{7}{57}\)

b: \(-\frac23-\left(\frac{x}{2}-75\%\right)=\left(\frac{3}{-4}-\frac98\right)^2:\frac{-3}{32}-1\frac13\)

=>\(-\frac23-0,5x+\frac34=\left(-\frac68-\frac98\right)^2\cdot\frac{-32}{3}-\frac43\)

=>\(-0,5x+\frac{1}{12}=\left(-\frac{15}{8}\right)^2\cdot\frac{-32}{3}-\frac43\)

=>\(-0,5x+\frac{1}{12}=\frac{225}{64}\cdot\frac{-32}{3}-\frac43=\frac{-75}{2}-\frac43=\frac{-225-8}{6}=\frac{-231}{6}\)

=>\(-0,5x=-\frac{231}{6}-\frac{1}{12}=\frac{-462-1}{12}=\frac{-463}{12}\)

=>\(x=\frac{463}{12}:0,5=\frac{463}{12}:\frac12=\frac{463}{12}\cdot2=\frac{463}{6}\)

sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

Cộng vế với vế 

\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4 

Vậy ta có đpcm 

1) Ta có 

\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)

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

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

2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)

a) \(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{2}{12}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{0}{6}=0\)

b) \(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{8}{20}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4-2-1}{10}=\dfrac{1}{10}\)

CHÚC EM HỌC TỐT NHÉ

\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

Ai trả lời đi please

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015