Câu hỏi của Lê Thị Minh Trang - Toán lớp 6 - Học toán với OnlineMath

Xem bài 1 nhé !

Bài 1:

Xét vế phải :

\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}\)\(-1=2\)\(\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left(\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\)

Đẳng thức được chứng tỏ là đúng

Bài 2 :

Đặt \(A'=\frac{3}{4}.\frac{4}{5}.\frac{7}{8}...\frac{4999}{5000}\)

Rõ ràng \(A< A'\)

SUY RA \(A^2< AA'=\frac{2}{50000}=\frac{1}{2500}=\left(\frac{1}{50}\right)^2\)

Nên \(A< \frac{1}{50}=0,02\)

Chúc bạn học tốt ( -_- )

Đặt \(M=\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{4999}{5000}\)

Xét: \(A.M=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}...\frac{4998}{4999}.\frac{4999}{5000}\)

\(\Leftrightarrow A.M=\frac{2.3.4.5...4998.4999}{3.4.5.6...4999.5000}\)

\(\Leftrightarrow A.M=\frac{2}{5000}\)

\(\Leftrightarrow A.M=\frac{1}{2500}\)

Mà \(0,02=\frac{1}{50}\)

\(\Rightarrow\frac{1}{2500}< \frac{1}{50}\)

\(\Rightarrow A.M< 0,02\)

\(\Rightarrow A< 0,02\)

Vậy A < 0,02.

2/3 > 0,02

4998/4999 > .... > 4/5 > 2/3 > 0,02

=> A = 2/3 . 4/5 ....4998/4999 .0,02

Vậy A > 2 

Đặt \(A'=\frac{3}{4}.\frac{5}{6}.\frac{7}{8}....\frac{4999}{5000}\)

Rõ ràng \(A< A'\)

\(\Rightarrow A^2< AA'A=\frac{2}{50000}=\frac{1}{2500}=\left(\frac{1}{50}\right)^2\)

Nên  \(A< \frac{1}{50}=0,02\)

\(A=\frac{1}{2}\times\frac{3}{4}......\frac{9999}{10000}\)

Đặt : \(B=\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}.......\frac{10000}{10001}\)

Vì \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};.....\frac{9999}{10000}< \frac{10000}{10001}\)

Nên A<B  mà A>0; B>0

\(\Rightarrow A^2< A\times B=\left(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}.....\frac{9999}{10000}\right)\times\left(\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}......\frac{10000}{10001}\right)\)\(=\frac{1}{2}\times\frac{2}{3}\times\frac{4}{5}......\frac{9999}{10000}\times\frac{10000}{10001}\)\(=\frac{1}{10001}< \frac{1}{10000}=\frac{1}{100^2}=0.01^2\)\(\Rightarrow A^2< 0.01^2\)hay A < 0.01

cách này mình tự nghĩ 

\(\hept{\begin{cases}A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\\B=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\end{cases}}\)

\(\Rightarrow A-B=\left(\frac{4}{7}-\frac{4}{7}\right)+\left(\frac{5}{7^3}-\frac{5}{7^3}\right)+\left(5-5\right)+\left(\frac{3}{7^2}-\frac{6}{7^2}\right)+\left(\frac{6}{7^4}-\frac{5}{7^4}\right)\)

\(\Rightarrow A-B=-\frac{3}{7^2}+\frac{1}{7^4}\)

\(\Rightarrow A-B=\frac{-3\times7^2}{7^4}+\frac{1}{7^4}\)

mà \(-3\times7^2< 1\Rightarrow\frac{1}{7^4}>\frac{-3\times7^2}{7^4}\Rightarrow B>A\)