Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
p=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)
=\(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\frac{50}{50}\)
=\(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
=\(50\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
p=50*S
\(\frac{S}{\text{p}}=\frac{1}{50}\)
Ta có: P = \(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{49}{1}\)
\(=\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{1}{49}\)
\(=\frac{50-1}{1}+\frac{50-2}{2}+\frac{50-3}{3}+...+\frac{50-49}{49}\)
\(=\frac{50}{1}-\frac{1}{1}+\frac{50}{2}-\frac{2}{2}+\frac{50}{3}-\frac{3}{3}+...+\frac{50}{49}-\frac{49}{49}\)
\(=\left(\frac{50}{1}+\frac{50}{2}+\frac{50}{3}+...+\frac{50}{49}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{49}{49}\right)\)
\(=50+50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)-49\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+1\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)}=\frac{1}{50}\)
Ta có:\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+....+\frac{48}{2}+\frac{49}{1}+50-50\)
\(=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\left(1+\frac{49}{2}\right)-50\)
\(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+....+\frac{50}{2}+\frac{50}{1}-50\)
\(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)+50-50\)
\(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)\)
mà \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}\)
\(=>\frac{S}{P}=\frac{1}{50}\)
Vậy \(\frac{S}{P}=\frac{1}{50}\)
Ta có: \(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
\(\Rightarrow P=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+1\)
\(\Rightarrow P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(\Rightarrow P=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)}=\frac{1}{50}\)
Vậy \(\frac{S}{P}=\frac{1}{50}\)
bài khó nhất nhé
2. Ta có :
\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
cộng vào 48 phân số đầu với 1, trừ phân số cuối đi 48 ta được :
\(P=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\left(\frac{49}{1}-48\right)\)
\(P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(P=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
\(P=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}=\frac{1}{50}\)
Bạn tham khảo Câu hỏi của Đoàn Phạm Hùng