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a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
\(a)15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\)
\(=15\frac{3}{13}-8\frac{3}{13}-3\frac{4}{7}\)
\(=7-3\frac{4}{7}\)
\(=6\frac{7}{7}-3\frac{4}{7}=3\frac{3}{7}\)
\(b)\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7\frac{4}{9}-4\frac{7}{11}-3\frac{4}{9}\)
\(=7\frac{4}{9}-3\frac{4}{9}-4\frac{7}{11}\)
\(=4-4\frac{7}{11}=\frac{7}{11}\)
2-2/19+2/43-2/1943
3-3/19+3/43-3/1943
=2(1-1/19+1/43-1/1943)
3(1-1/19+1/43-1/1943)
=2/3
Chúc bạn học tốt!!!!!
ta có :
\(\frac{1}{2.3}>\frac{1}{3^2}>\frac{1}{4.3};\frac{1}{3.4}>\frac{1}{4^2}>\frac{1}{4.5}....\)
Tương tự ta sẽ có :
\(\frac{1}{2^2}+\frac{1}{2.3}+.+\frac{1}{99.100}>A>\frac{1}{2^2}+\frac{1}{3.4}+..+\frac{1}{100.101}\)
hay ta có :
\(\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}>A>\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{100}-\frac{1}{101}\)
hay \(\frac{1}{4}+\frac{1}{2}-\frac{1}{100}>A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\)
hay ta có : \(\frac{1}{4}+\frac{1}{2}>A>\frac{1}{4}+\frac{1}{3}-\frac{31}{300}\Leftrightarrow\frac{3}{4}>A>\frac{12}{25}\)
vậy ta có điều phải chứng minh
A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
A = \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)+ \(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)
A = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)\(+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)\(+\frac{1}{9}-\frac{1}{10}\)
A = \(\frac{1}{5}-\frac{1}{10}\)
A = \(\frac{1}{10}\)
ta có :1/30+1/42+1/56+1/72+1/90=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10=1/5-1/6+1/6-1/7=1/7-1/8+1/8-1/9+1/9-1/10=1/5-1/10=1/10
a)\(\frac{24}{146}=\frac{24.13}{146.13}=\frac{312}{1898}\)
\(\frac{6}{13}=\frac{6.146}{13.146}=\frac{876}{1898}\)
b) \(\frac{7}{30}=\frac{7.60.40}{30.60.40}=\frac{16800}{72000}\)
\(\frac{13}{60}=\frac{13.30.40}{60.30.40}=\frac{15600}{72000}\)
\(\frac{-9}{40}=\frac{-9.30.60}{40.30.60}=\frac{-16200}{72000}\)
c)\(\frac{17}{60}=\frac{17.18.90}{60.18.90}=\frac{27540}{97200}\)
\(\frac{-5}{18}=\frac{5.60.90}{18.60.90}=\frac{27000}{97200}\)
\(\frac{-64}{90}=\frac{-64.18.60}{90.18.60}=\frac{-69120}{97200}\)
Mệt quá đi
Câu 1:
\(S=\frac{10}{7}+\frac{10}{7^2}+\frac{10}{7^3}+...+\frac{10}{7^{10}}\)
\(\frac{1}{7}S=\frac{10}{7^2}+\frac{10}{7^3}+....+\frac{10}{7^{11}}\)
\(\rightarrow\)\(\left(1-\frac{1}{7}\right).S=\frac{10}{7}-\frac{10}{7^{11}}\)
=> \(S=\frac{10.7^{10}-10}{7^{10}.6}\)
5,4999055
5,4999055