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\(S=\dfrac{27+4500+135+550.2}{2+4+6+...+14+16+18}\)
\(S=\dfrac{4527+135+1100}{\left(2+18\right).9.2}\)
\(S=\dfrac{4662+1100}{20.9.2}\)
\(S=\dfrac{5762}{90}\)
\(S=\dfrac{2881}{45}\)
`S=(27+4500+135+550.2)/(2+4+6+…+14+16+18)`
`S=(27+4500+135+1100)/((2+18)+(4+16)+…+(8+12)+10)`
`S=((27+135)+(4500+1100))/(20+20+…+20+10)` `(`Có `4` số hạng `20)`
`S=(162+5600)/(20.4+10)`
`S=5762/90`
`S=2881/45`
\(\frac{27+4500+135+550.2}{2+4+6+...+16+18}\)= \(\frac{27+4500+135+1100}{\left(18+2\right).\left[\left(18-2\right):2+1\right]:2}\)= \(\frac{5762}{20.19:2}\)=\(\frac{5762}{190}\)=\(\frac{2881}{95}\)
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Bài làm
Ta có:
\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)
=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)
hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)
=> \(S=\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)
Do đó: \(S=\frac{1}{2}\)
# Chúc bạn học tốt #
#)Giải :
Câu 1 :
Đặt \(A=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{27}\)
\(\Rightarrow A>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}\)( 8 số hạng )
\(\Rightarrow A>\frac{8}{27}=\frac{8}{27}\)
\(\Rightarrow A>\frac{8}{27}\)
#~Will~be~Pens~#
Câu 1:(trội)
Ta có:\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}=\frac{8}{27}\left(đpcm\right)\)
Câu 2:\(D=\frac{2^{25}.3^{15}+3^{15}.5.2^{26}}{2^{25}.3^{17}+3^{15}.2^{25}}=\frac{2^{25}3^{15}\left(1+5.2\right)}{2^{25}3^{15}\left(3^2+1\right)}=\frac{11}{10}\)
a) 70+140+ 77 có chia hết cho 7 vì 70 chia hết cho 7 ; 140 chia hết cho 7 ; 77 chia hết cho 7
b) 14+ 27+ 77 không chia hết cho 7 vì 14 chia hết cho 7;27 không chia hết cho 7; 77 chia hết cho 7
a) 70+140+77
=7.10+7.20+7.11
=7.(10+20+7)
=7.37\(⋮7\)
b)14+27+77
=2.7+27+7.11
=7.(2+11)+27
=7.13+27 \(⋮\)7
\(S=\frac{27+4500+135+550.2}{2+4+6+...+14+16+18}\)
\(S=\frac{4527+135+1100}{\left(2+18\right).9:2}=\frac{4662+1100}{20.9:2}=\frac{5762}{90}=\frac{2881}{45}\)
\(B=\frac{27+4500+135+1100}{\frac{\left(2+18\right).9}{2}}=\frac{5762}{10.9}=\frac{2881}{45}\)