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

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

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

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

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

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

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

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

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

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

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

 = \(\frac{4862}{6561}\)

KẾT QUẢ BẰNG \(\frac{4862}{6561}\)

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

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

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

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

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

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

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

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

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

Cái này ko làm theo quy tắc gì hết em nhé, chỉ là cách làm của dạng này thôi nha !!!!!

( nhớ ra nhiều bài để giải kiếm sp nha chứ dạo này ko lm đc j hết )

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.

C= (1.3.5.....199)/(2.4.6.....200)

=> C^2= (1^2. 3^2. 5^2......199^2)/(2^2. 4^2. 6^2......200^2)

Ta có k^2 > k^-1 = (k-1)(k+1) nên 2^2 > 1.3

                                                 4^2 > 3.5 

                                                 ....

                                                 200^2 > 199.201

=> C^2 < (1^2.3^2.5^2.....199^2) / (1.3)(3.5)(5.7).....(199.201)

ta có:  (1^2.3^2.5^2.....199^2) / (1.3)(3.5)(5.7).....(199.201) 

=1/201

Do đó C^2 <1/201

Vậy C^2 < 1/201

 Ta có : ^{\(C=\frac{1}{2}\times\frac{3}{4}\times.....\times\frac{199}{200}\)}

      \(\Rightarrow C< \frac{2}{3}\times\frac{4}{5}\times.......\times\frac{200}{201}\)

      \(\Rightarrow C^2< \frac{2}{3}\times\frac{4}{5}\times......\times\frac{200}{201}\times\frac{1}{2}\times\frac{3}{4}\times.....\times\frac{199}{200}\)

     \(\Rightarrow C^2< \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times........\times\frac{199}{200}\times\frac{200}{201}\)

     \(\Rightarrow C^2< \frac{1}{201}\left(đ.p.c.m\right)\)

a)\(5^{2x-3}-2\cdot5^2=5^2\cdot3\)

\(5^{2x-3}-2\cdot25=75\)

\(5^{2x-3}-50=75\)

\(5^{2x-3}=125\)

\(125=5^3\)

\(5^3=5^{\left(3+3\right):2}=5^3\Rightarrow x=3\)

Vậy \(x=3\)

b)\(\frac{2}{9}\cdot\left(5x+1\right):2-\frac{1}{18}=\frac{5}{36}\)

\(=\frac{2}{9}\cdot\left(5x+1\right):2=\frac{5}{36}+\frac{1}{18}\)

\(=\frac{2}{9}\cdot\left(5x+1\right)=\frac{7}{36}\cdot2\)

\(5x+1=\frac{7}{18}:\frac{2}{9}\)

\(x=\left(\frac{7}{4}-1\right):5=\frac{3}{20}\)

\(\Rightarrow x=\frac{3}{20}\)

a, \(5^{2x-3}-2.5^2=5^2.3\)

\(\Leftrightarrow5^{2x-3}-2.25=25.3\)

\(\Leftrightarrow5^{2x-3}-50=75\)

\(\Leftrightarrow5^{2x-3}=125\)

\(\Leftrightarrow5^{2x-3}=5^3\)

\(\Leftrightarrow2x-3=3\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\frac{2}{9}.\left(5x+1\right):2-\frac{1}{18}=\frac{5}{36}\)

\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right):2=\frac{7}{36}\)

\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right)=\frac{7}{36}.2\)

\(\Leftrightarrow\frac{2}{9}.\left(5x+1\right)=\frac{7}{18}\)

\(\Leftrightarrow5x+1=\frac{7}{18}:\frac{2}{9}\)

\(\Leftrightarrow5x+1=\frac{7}{18}.\frac{9}{2}\)

\(\Leftrightarrow5x+1=\frac{7}{4}\)

\(\Leftrightarrow5x=\frac{3}{4}\)

\(\Leftrightarrow x=\frac{3}{4}:5=\frac{3}{4}.\frac{1}{5}=\frac{3}{20}\)