- Tính tổng M=\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+...+\(\frac{1}{97.99}\)
- Cho S=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\)và P=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\). Tính\(\frac{S}{P}\)
- Tìm x biết \(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
- Không quy đồng mẫu hãy so sánh A=\(\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}\); B=\(\frac{-9}{10^{2011}}+\frac{-19}{10^{2011}}\)
- Cho phân số \(\frac{13}{18}\). Nếu thêm 52 vào tử thì phải thêm vào mẫu bao nhiêu để giá trị phân số không thay đổi
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
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
\(\frac{S}{\text{p}}=\frac{1}{50}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)
S=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\frac{98}{99}\)
S=\(\frac{49}{99}\)
S = 1/3.5 + 1/5.7 + 1/7.9 +...+1/97.99
S = 1 - ( 1/3+1/3-1/5+1/5-+1/7+1/7+...+1/97-1/99)
S = 1 - 1/99
S = 98/99
\(2S=\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{97.99}\)
\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(2S=\frac{1}{3}-\frac{1}{99}\)
\(2S=\frac{33}{99}-\frac{1}{99}\)
\(S=\frac{32}{99}:2\)
\(S=\frac{16}{99}\)
Ta có: P = \(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{49}{1}\)
\(=\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{1}{49}\)
\(=\frac{50-1}{1}+\frac{50-2}{2}+\frac{50-3}{3}+...+\frac{50-49}{49}\)
\(=\frac{50}{1}-\frac{1}{1}+\frac{50}{2}-\frac{2}{2}+\frac{50}{3}-\frac{3}{3}+...+\frac{50}{49}-\frac{49}{49}\)
\(=\left(\frac{50}{1}+\frac{50}{2}+\frac{50}{3}+...+\frac{50}{49}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{49}{49}\right)\)
\(=50+50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)-49\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+1\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)}=\frac{1}{50}\)
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{12}{60}+\frac{-5}{60}\)
\(=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{2}{3}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
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 :
\(P=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\left(\frac{49}{1}-48\right)\)
\(P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(P=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
\(P=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}=\frac{1}{50}\)
câu 5đáp án là72