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\(C=\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\right)\)
\(\Rightarrow\frac{2}{3}C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{47\cdot49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{49}\)
\(\Rightarrow\frac{2}{3}C=\frac{46}{147}\)
\(\Rightarrow C=\frac{46}{147}:\frac{2}{3}\)
\(\Rightarrow C=\frac{23}{49}\)
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
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=\(\frac{4}{13}\)
\(C=\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{47.49}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{47}-\frac{1}{49}\)
\(=\frac{1}{3}-\frac{1}{49}\)
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 = \(\frac{1}{4}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
1/5.7 + 1/7.9 + 1/9.11 + ... + 1/49.51
= 1/2 . (2/5.7 + 2/7.9 + 2/9.11 + ... + 2/49.51)
= 1/2 . (1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/49 - 1/51)
= 1/2 . (1/5 - 1/51)
= 1/2 . 46/255
= 23/255
S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...\frac{1}{49}-\frac{1}{51}\)
S = \(\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))
M= 3.(\(\frac{1}{5}-\frac{1}{61}\))
M=\(\frac{168}{305}\)
\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(M=\frac{84}{305}\)
a ) \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)
\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{203}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{203}\right)\)
\(=\frac{3}{2}.\left(\frac{203}{1015}-\frac{5}{1015}\right)\)
\(=\frac{3}{2}.\frac{198}{1015}\)
\(=\frac{297}{1015}\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Tham khảo nha !!!
\(M=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{61.63}\)
\(2M=2.\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{61.63}\right)\)
\(2M=3.\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{61.63}\right)\)
\(2M=3.\left(\frac{1}{5}-\frac{1}{63}\right)\)
\(2M=\frac{3.58}{315}=\frac{58}{105}\)
\(M=\frac{58}{105}.\frac{1}{2}=\frac{29}{105}\)
Ta có thể vt gọn thành :
M = \(\frac{3}{2}\).( \(\frac{1}{5}\)\(-\)\(\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\)... \(+\frac{1}{61}-\frac{1}{63}\))
M = \(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{63}\right)\)
M = \(\frac{3}{2}.\frac{58}{315}\)
M = \(\frac{29}{105}\)