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\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}\right)+\frac{1}{100}.\left(\frac{1}{2}-\frac{1}{102}\right)+\frac{1}{100}.\left(\frac{1}{3}-\frac{1}{103}\right)+...+\frac{1}{100}.\left(\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)\)
\(A=\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}\right)+\frac{1}{25}.\left(\frac{1}{2}-\frac{1}{27}\right)+\frac{1}{25}.\left(\frac{1}{3}-\frac{1}{28}\right)+...+\frac{1}{25}.\left(\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)\)
\(B=\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)\)\(B=\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)\)
Ta thấy biểu thức trong ngoặc của hai vế A và B giống nhau
Vậy A : B = \(\frac{1}{100}:\frac{1}{25}=\frac{1}{4}\)
\(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
\(\Rightarrow A=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{24.25}\right)+\left(\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{124.125}\right)\)
\(A=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{24}-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+...+\frac{1}{124}-\frac{1}{125}\right)\)
\(A=\left(1-\frac{1}{25}\right)+\left(\frac{1}{101}-\frac{1}{125}\right)\)
\(A=\frac{24}{25}+\frac{24}{12625}\)
Bạn tự tính luôn nha trog máy tính của mình là : 0,961... ( k làm thành phân số được )
Ta có : \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=\frac{81}{10}\)
A. = 1/2-1/3+1/3-1/4+1/4-1/5...+1/101-1/102=1/2-1/102=25/51.
B. =1/5-1/10+1/10-1/15+...+1/115-1/120=1/5-1/120=23/120.
C. = 1/5-1/7+1/7-1/9+1/9-1/11+...+1/997-1/999=1/5-1/999=994/4995.
Minh kiem tra bang may tinh roi do.
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}\)
\(=1-\frac{1}{102}\)
\(=\frac{101}{102}\)
a) MC :24
\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{1\times8+3\times3-7\times2}{24}=\frac{3}{24}=\frac{1}{8}\)
b)MC : 56
\(\frac{3}{14}+\frac{5}{8}-\frac{1}{2}=\frac{3\times4+5\times7-1\times28}{56}=\frac{19}{56}\)
c) MC: 36
\(\frac{1}{4}-\frac{2}{3}-\frac{11}{18}=\frac{1\times9-2\times12-11\times2}{36}=\frac{-37}{36}\)
d) MC: 312
\(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}=\frac{1\times78+5\times26-1\times24-7\times39}{312}=\frac{-89}{312}\)
tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
B = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{63}\)
B = \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)+\frac{1}{63}\)
B = \(1+\frac{1}{5}+\frac{3}{40}+\frac{1}{63}\)
B = \(1\frac{11}{40}+\frac{1}{63}\)
B = \(1\frac{733}{2520}\)
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A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
299.A= 299.(\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\))
299.A=\(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}=\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
A= \(=\frac{1}{299}\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Tương tự
B=\(\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Hai dấu ngoặc ở biểu thức A và biểu thức B như nhau
Vậy \(A:B=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)