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=19/25.3/7+19/25.25/19-3/7.19/25-3/7.7/3
=19/25.3/7+1-3/7.19/25-1
=19/25.3/7-19/25+1-1
=0
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ta có:
\(\frac{1}{19}>\frac{1}{40};\frac{1}{29}>\frac{1}{40};\frac{1}{31}>\frac{1}{40};\frac{1}{39}>\frac{1}{40}\)
\(\Rightarrow\frac{1}{19}+\frac{1}{29}+\frac{1}{31}+\frac{1}{39}>\frac{1}{40}.4\)=\(\frac{1}{10}\)
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Xin phép đc ghi lại đề như sau \(C=\frac{19}{25}\left(\frac{3}{7}+\frac{25}{19}\right)-\frac{3}{7}\left(\frac{19}{25}-\frac{7}{3}\right)\)\(=\frac{19}{25}.\frac{3}{7}+\frac{19}{25}.\frac{25}{19}-\frac{3}{7}.\frac{19}{25}+\frac{3}{7}.\frac{7}{3}=\frac{19}{25}.\frac{3}{7}-\frac{3}{7}.\frac{19}{25}+\frac{19}{25}.\frac{25}{19}+\frac{3}{7}.\frac{7}{3}=0+1+1=2\)
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\(B=\dfrac{20^{19}+1}{20^{20}+1}< \dfrac{20^{19}+1+19}{20^{20}+1+19}=\dfrac{20^{19}+20}{20^{20}+20}\)
\(B< \dfrac{20.\left(20^{18}+1\right)}{20.\left(20^{19}+1\right)}\)
\(B< \dfrac{20^{18}+1}{20^{19}+1}\)
\(B< A\)
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Gọi số tự nhiên cần tìm là x.
Đặt A=x-5 x chia 29 dư 5
=> A chia hết cho 29 x chia 31 dư 28
=> A chia 31 dư 23 =>A=31k+23
Cho k=0,1,2,3,... ta thấy khi k=3 thì A=116 chia hết cho 29
Vậy x=A+5=116+5=121.
\(\frac{19^{31}}{19^{30}}=\frac{19^{30}.19}{19^{30}}=19\)
\(\frac{19^{32}}{19^{31}}=\frac{19^{31}.19}{19^{31}}=19\)
\(\Rightarrow\frac{19^{31}}{19^{30}}=\frac{19^{32}}{19^{31}}\)