\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)

\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)

\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)

Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
 

Lam mô a Di Đà Phật 

 = \(\frac{4862}{6561}\)

KẾT QUẢ BẰNG \(\frac{4862}{6561}\)

?

39.337+64.337=337.103=337.100+337.3=33700+1011=34711

Kiến thức cần nhớ:

Đấy là dạng tính nhanh phân số mà mẫu nọ gấp một số lần mẫu kia, ta nhân cả hai vế với số lần, trừ vế cho vế, triệt tiêu các hạng tử giống nhau, rút gọn ta được tổng cần tìm.

A                =     \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\) + \(\dfrac{1}{54}\)+...+ \(\dfrac{1}{1458}\)+\(\dfrac{1}{4374}\)

A \(\times\) 3         = \(\dfrac{3}{2}\)+\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\) + \(\dfrac{1}{54}\)+...+ \(\dfrac{1}{1458}\)

A \(\times\) 3 - A    =  \(\dfrac{3}{2}\) - \(\dfrac{1}{4374}\)

A \(\times\) ( 3  - 1) =  \(\dfrac{6561}{4374}\) - \(\dfrac{1}{4374}\)

A \(\times\) 2           =  \(\dfrac{6560}{4374}\)

A  \(\times\) 2          = \(\dfrac{3280}{2187}\)

A                  = \(\dfrac{3280}{2187}\): 2

A                  = \(\dfrac{1640}{2187}\)

????????????

Đặt S =\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\)

3S = \(3\times\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

3S \(=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\)

3S - S \(=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

2S = \(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{1458}-\frac{1}{4374}\)

2S = \(\frac{3}{2}-\frac{1}{4374}\)

2S = \(\frac{3280}{2187}\)

\(\Rightarrow S=\frac{3280}{2187}:2=\frac{4373}{8748}\)

Đáp án cuối cùng của "Ông nội bây" sai rùi phải là :

=>  \(s=\frac{3280}{2187}:2=\frac{3280}{4374}\)

Còn lại đúng hết nên mk sẽ cho bn  3 h

Ta có: \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+...+\frac{1}{1458}+\frac{1}{4374}\)

\(\Leftrightarrow3\cdot C=3\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

\(\Leftrightarrow3\cdot C=\frac{3}{2}+\frac{3}{6}+\frac{3}{18}+\frac{3}{54}+...+\frac{3}{1458}+\frac{3}{4374}\)

\(\Leftrightarrow3\cdot C-C=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{486}+\frac{1}{1458}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

\(\Leftrightarrow2\cdot C=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{486}+\frac{1}{1458}-\frac{1}{2}-\frac{1}{6}-\frac{1}{18}-\frac{1}{54}-...-\frac{1}{4374}\)

\(\Leftrightarrow2\cdot C=\frac{3}{2}-\frac{1}{4374}\)

\(\Leftrightarrow2\cdot C=\frac{6561}{4374}-\frac{1}{4374}=\frac{3280}{2187}\)

\(\Leftrightarrow C=\frac{3280}{2187}:2=\frac{3280}{2187}\cdot\frac{1}{2}=\frac{1640}{2187}\)

anh Nguyễn Lê Phước Thịnh ra nhiều cuộc thi hơn đc ko ạ, mong anh giúp ạ ( lớp 7 nha anh )

Ta thấy:

\(P=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4374}\\ =\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\right)\\ =\frac{1}{2}\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

Mà:

\(\frac{1}{3}P=\frac{1}{2}\cdot\frac{1}{3}\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^7}\right)\\ =\frac{1}{2}\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

Suy ra: \(P-\frac{1}{3}P=\frac{1}{2}\left[\left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\right]\)

hay \(\frac{2}{3}P=\frac{1}{2}\left(\frac{1}{3^0}-\frac{1}{3^8}\right)=\frac{1}{2}\left(1-\frac{1}{6561}\right)=\frac{3280}{6561}\)

Vậy \(P=\frac{3280}{6561}:\frac{2}{3}=\frac{1640}{2187}\).

Chúc bạn học tốt nha.

A) bạn xem lại đề ạ

B) 1/2 + 1/6 + 1/ 12 + 1/ 20 + ...+ 1/ 9900
=1/2+1/6+1/12+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100
=99/100
C) Biến đổi tử số và mẫu số ta có

- Tử số: 20,2 x 5,1 - 30,3 x 3,4 + 14,58
          = 103,02 - 103,02 + 14,58
          = 14,58
- Mẫu số: 14,58 x 460 + 7,29 x 540 x 2
          = 14,58 x 460 + 14,58 x 540
          = 14,58 x (460 + 540)
          = 14,58 x 1000
          = 14580

Thay vào ta có: = 14,58 : 14580
                         = 0,001
Vậy 20.2*5.1-30.3*3.4+14.56/ 14.58*460+7.29 *540*2 = 0,001.

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}\)

\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)

Gọi  \(B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(\Rightarrow\frac{1}{3}B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(\Rightarrow B-\frac{1}{3}B=1-\frac{1}{243}\)

\(\Rightarrow\frac{2}{3}B=\frac{242}{243}\)

\(\Rightarrow B=\frac{121}{81}\)

Suy ra  \(A=\frac{1}{2}B=\frac{1}{2}.\frac{121}{81}=\frac{121}{162}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)