\(\frac{3}{2^2}.\frac{2^3}{3^2}.\frac{5.3}{4^2}.......\frac{3.3333}{100^2}\)

mk ko nghĩ ra phần sau chắc là rút gọn thì phải!!?

bạn làm thử xem

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)

\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)

\(x=\frac{1}{100}.\frac{101}{2}\)

\(x=\frac{101}{200}\)

\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)

\(X=\frac{1}{100}.\frac{101}{2}\)

\(X=\frac{101}{200}\)

Study well 

A=1.3/2^2.2.4/3^2.3.5/4^2...99.101/100.100

A=(1.2.3...99/2.3.4...100).(3.4.5...101/2.3.4...100)

A=1/100.101/2

A=101/200

200.A=200.101/200

200.A=101

P=1.3/2.2 . 2.4/3.3 . 3.5/4.4 ... . 99.101/100.100

P=1.2.3....99/2.3.4...100 . 3.4.5...101/2.3.4...100

P=1/100 . 101/2

P=101/200

p=101/200

P = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

P = \(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)

P = \(\frac{1}{100}.\frac{101}{2}\)

P = \(\frac{101}{200}\)

\(M=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot....\cdot\frac{99\cdot101}{100\cdot100}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\frac{1\cdot101}{2\cdot100}=\frac{101}{200}\)Vậy M = \(\frac{101}{200}\)

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

\(M=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

                                                 \(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)

                                             \(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)

                                              \(=\frac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)

                                                  \(=\frac{1.2.3.....99}{2.3.4.....100}.\frac{3.4.5.....101}{2.3.4.....100}\)

                                                    \(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

                                              Ủng hộ mk nha,chúc bn học tốt!!!

bn ơi mjk chưa học số có mũ bn giúp mjk nốt nha

Ta có :

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99\)\(\left(1\right)\)

gọi B là biểu thức trong ngoặc

Lại có :

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(B< 1-\frac{1}{100}< 1\)

\(\Rightarrow A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(1-\frac{1}{100}\right)>98\)

\(\Rightarrow A>98\)\(\left(2\right)\)

từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(98< A< 99\)

vậy A không phải là số tự nhiên

phần bạn đánh dấu (1) thì A<99 vì A= 99 trừ đi một số mà