\(\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3.3}< \frac{1}{2.3}\)

......

\(\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha

B<3\4 là đúng

khó thế

Sửa đề: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

Ta có: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(3a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}\)

=>\(3a+a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

Đặt \(b=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(3b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}\)

=>\(3b+b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(4b=-1-\frac{1}{3^{2023}}=\frac{-3^{2023}-1}{3^{2023}}\)

=>\(b=\frac{-3^{2023}-1}{4\cdot3^{2023}}\)

Ta có: \(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1+\frac{-3^{2023}-1}{4\cdot3^{2023}}-\frac{2024}{3^{2024}}=1+\frac{-3^{2024}-3}{4\cdot3^{2024}}-\frac{8096}{4\cdot3^{2024}}\)

=>\(4a=1-\frac{3^{2024}+8099}{4\cdot3^{2024}}=1-\frac14-\frac{8099}{4\cdot3^{2024}}=\frac34-\frac{8099}{4\cdot3^{2024}}\)

=>\(4a<\frac34\)

=>\(a<\frac{3}{16}\)

mà \(\frac{3}{16}<1<\frac{20}{3}\)

nên \(a<\frac{20}{3}\)

bằng 2 nha bạn hải nam

gải ra hộ tớ

Ta có :

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)

\(=\)\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{18}{19}.\frac{19}{20}\)

\(=\)\(\frac{1.2.3.....18.19}{2.3.4.....19.20}\)

\(=\)\(\frac{1}{20}\)

Vì \(\frac{1}{20}>\frac{1}{21}\)nên \(A>\frac{1}{21}\)

Vậy \(A>\frac{1}{21}\)

a)  - 2 3 .125.32. ( − 76 ) > 0 12.74 . - 3 4 . ( − 395 ) < 0 ⇒ ( − 2 ) 3 .125.32. ( − 76 ) > 12.74. ( − 3 ) 4 . ( − 395 )

b)  ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > 0 ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 ) < 0 ⇒ ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 )

Ta thấy:

\(\frac{1}{2^2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}<\frac{1}{2.3}\)

................

\(\frac{1}{19^2}<\frac{1}{18.19}\)

Cộng vế với vế ta có:

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{19^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{18.19}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}\)\(=1-\frac{1}{19}=\frac{18}{19}>\frac{18}{40}=\frac{9}{20}\)

Kết luận: ....>.....

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B

Xét 3 số TN liên tiếp \(\left(n-1\right);n;\left(n+1\right)\) ta có

\(\left(n-1\right).n.\left(n+1\right)=n.\left(n^2-1\right)=n^3-n< n^3\)

\(\Rightarrow A\le\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{20.21.22}=\)

\(=\dfrac{1}{2}\left(\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{22-20}{20.21.22}\right)=\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{20.21}-\dfrac{1}{21.22}\right)=\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{21.22}\right)=\dfrac{1}{2^2}-\dfrac{1}{2.21.22}< \dfrac{1}{2^2}\)