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\(Q=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{47.49}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(=\frac{1}{3}-\frac{1}{49}\)
\(=\frac{46}{147}\)
Vậy \(Q=\frac{46}{147}\)
Ta có : \(\frac{2}{3}Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{47.49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{49}=\frac{49}{147}-\frac{3}{147}=\frac{46}{147}\)
\(\Rightarrow Q=\frac{46}{147}\div\frac{2}{3}=\frac{138}{294}=\frac{23}{49}\)
Vậy ...
\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}\)
\(A=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot....\cdot\frac{2499}{50^2}\)
\(A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{49\cdot51}{50\cdot50}\)
\(A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(A=\frac{1\cdot51}{2\cdot50}=\frac{51}{100}\)
\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\cdot\frac{2499}{2500}\)
\(\Rightarrow A=\frac{1.3}{2\cdot2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{49.51}{50.50}\)
\(\Rightarrow A=\frac{1.3.2.4.3.5.....49.51}{2.2.3.3.4.4.....50.50}\)
\(\Rightarrow A=\frac{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot49\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot51\right)}{\left(3\cdot4\cdot\cdot\cdot\cdot\cdot50\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot50\right)}\)
\(\Rightarrow A=\frac{1.51}{2\cdot50}\)
\(\Rightarrow A=\frac{51}{100}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
Gọi 2/3.5 +2/5.7 +2/7.9 +...+2/97.99 là A
A=2/3.5 +2/5.7 +2/7.9+...+ 2/97.99
A= 1.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
A=1.(1/3-1/99)
A=1.32/99
A=32/99
Ta có: A>8/25
=>32/99>8.25
Vậy 2/3.5+2/5.7+2/7.9+...+2/97.99>8/25
k cho mk nha!!!
\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
\(C=\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{47.49}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{47}-\frac{1}{49}\)
\(=\frac{1}{3}-\frac{1}{49}\)
a)
C = \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+........+\frac{3}{47.49}\)
C = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-.........-\frac{1}{47}+\frac{1}{47}-\frac{1}{49}\)
C = \(\frac{1}{3}-\frac{1}{49}\)
C = \(\frac{49}{147}-\frac{3}{147}\)
C = \(\frac{46}{147}\)
b) \(\frac{7}{2}.\left(\frac{1}{2}-x\right)-\frac{1}{8}=\frac{3}{4}\)
\(\frac{7}{2}.\left(\frac{1}{2}-x\right)=\frac{3}{4}+\frac{1}{8}\)
\(\frac{7}{2}.\left(\frac{1}{2}-x\right)=\frac{24}{32}+\frac{4}{32}\)
\(\frac{7}{2}.\left(\frac{1}{2}-x\right)=\frac{28}{32}\)
\(\frac{1}{2}-x=\frac{28}{32}:\frac{7}{2}\)
\(\frac{1}{2}-x=\frac{7}{8}.\frac{2}{7}\)
\(\frac{1}{2}-x=\frac{1}{4}\)
\(x=\frac{1}{2}-\frac{1}{4}\)
\(x=\frac{2}{4}-\frac{1}{4}=\frac{1}{4}\)
Vậy x = \(\frac{1}{4}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ..... + \(\frac{1}{97}\)- \(\frac{1}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{99}\))
= 2 . \(\frac{2}{3}\)
= \(\frac{4}{3}\)
32% = \(\frac{32}{100}\)= \(\frac{8}{25}\)
\(\frac{4}{3}\)> \(\frac{8}{25}\)=> \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\)+ ..... + \(\frac{2}{97.99}\)> 32%
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}=\frac{800}{2475}\)
\(32\%=\frac{8}{25}=\frac{792}{2475}\)
\(\frac{800}{2475}>\frac{792}{2475}\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}>32\%\)
\(C=\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\right)\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{46}{147}\)
\(\Rightarrow C=\frac{46}{147}:\frac{2}{3}\)
\(\Rightarrow C=\frac{23}{49}\)
3/3.5+3/5.7+3/7.9+.....+3/47.49
=1-1/5+1/5-1/7+...+1/47-1/49
=1-1/49
=48/49