Câu 1: Chứng tỏ rằng:
\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{26}+\frac{1}{27}>\frac{8}{27}\)
Câu 2: Tính:
\(D=\frac{2^{10}.6^{15}+3^{14}.15.4^{13}}{2^{18}.18^7.3^3+3^{15}.2^{25}}\)
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a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
A) \(\frac{2}{3}+\frac{1}{5}-\frac{10}{7}=\frac{13}{15}-\frac{10}{7}=\frac{-59}{105}\)
B) \(\frac{7}{12}-\frac{27}{18}.\frac{2}{18}=\frac{7}{12}-\frac{3}{2}.\frac{2}{18}=\frac{7}{12}-\frac{1}{6}=\frac{5}{12}\)
C) \(\left(\frac{23}{41}-\frac{15}{82}\right).\frac{41}{25}=\frac{31}{82}.\frac{41}{25}=\frac{31}{50}\)
D) \(\left(\frac{4}{5}+\frac{1}{2}\right).\left(\frac{3}{13}-\frac{8}{13}\right)=\frac{13}{10}.\frac{-5}{13}=\frac{-1}{2}\)
a, = \(\frac{-59}{105}\)
b = \(\frac{7}{144}\)
c= \(\frac{31}{50}\)
d=\(\frac{-1}{2}\)
#)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:\(D=\frac{2^{25}.3^{15}+3^{15}.5.2^{26}}{2^{25}.3^{17}+3^{15}.2^{25}}=\frac{2^{25}3^{15}\left(1+5.2\right)}{2^{25}3^{15}\left(3^2+1\right)}=\frac{11}{10}\)