Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
Bn tự tính típ nha
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)