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\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656
A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)
A=202202.5/2424
A=417/1/12
A=202202.1/1212+202202.1/2020+202202.1/3030+202202.1/4242+202202.1/5656
A=202202.(1/1212+1/2020+1/3030+1/4242+1/5656)
A=202202.5/2424
A=5005/12
a,3^200 và 2^300
3^200=(3^2)^100=9^100
2^300=(2^3)^100=8^100
Vì 9^100>8^100=>3^200>2^300
Vậy 3^200>2^300
b, 71^50 và 37^75
71^50=(71^2)^25=5041^25
37^75=(37^3)^25=50653^25
Vì 5041^25<50653^25=> 71^50<37^75
Vậy 71^50<37^75
c, 201201/202202 và 201201201/202202202
201201201/202202202=201201/202202
=> 201201/202202=201201201/202202202
Vậy 201201/202202=201201201/202202202
a)
Ta có:3200=32.100=(32)100=9100
2300=23.100=(23)100=8100
Vì 9100>8100
Nên 3200>2300
b)
Ta có: 7150=712.25=(712)25=504125
3775=373.25=(373)25=5065325
Vì 504125<5065325
Nên 7150<3775
c)
Ta có:
201201/202202=201.1001/202.1001=201/202
201201201/202202202=201.1001001/202.1001001001= 201/202
Vì 201/202=201/202
Nên 201201/202202=201201201/202202202
a. 3200 = (32)100 = 9100
2300 = (23)100 = 8100
Vì 9100 > 8100 => 3200 > 2300
a)\(\frac{-15}{90}=\frac{-1}{6}\)=\(\frac{-5}{30}\);\(\frac{120}{600}=\frac{1}{5}\)=;\(\frac{-75}{150}=\frac{-1}{2}\)
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Bạn cộng các mẫu trong hoặc và giữ nguyên tử nếu kết quả trong hoặc rút gọn đc thì rút luôn. Đây là cách làm trong hoặc. Tính trong hoặc xong bạn chỉ việc nhân lại với nhau thôi, kết quả cuối cùng rút đc thì rút luôn( ko đc thì thôi, đừng cố rút gọn)
\(=\frac{2002}{12}+\frac{2002}{20}+\frac{2002}{30}+\frac{2002}{42}+\frac{2002}{56}\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)\)
\(=2002.\frac{5}{24}\)
\(=\frac{5005}{12}\)
5005/12
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