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a) \(2\frac{3}{13}-\frac{5}{9}-\left(\frac{3}{13}+\frac{4}{9}\right)\)
= \(\frac{29}{13}-\frac{5}{9}-\left(\frac{3}{13}+\frac{4}{9}\right)\)
= \(\left(\frac{29}{13}-\frac{3}{13}\right)-\left(\frac{5}{9}+\frac{4}{9}\right)\)
= \(2-1\)
= \(1\)
b) \(17\frac{4}{16}+\frac{3}{4}-\left(2\frac{3}{12}+75\%\right)\)
= \(\frac{69}{4}+\frac{3}{4}-\left(\frac{27}{12}+\frac{3}{4}\right)\)
= \(\left(\frac{69}{4}+\frac{3}{4}\right)-\left(\frac{27}{12}+\frac{3}{4}\right)\)
= \(18-3\)
= \(15\)
c) \(\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+....+\frac{6}{101.103}+\frac{6}{103.106}\)
= \(3.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+....+\frac{2}{101.103}+\frac{2}{103.106}\right)\)
= \(3.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{103}+\frac{1}{103}-\frac{1}{106}\right)\)
= \(3.\left(\frac{1}{5}-\frac{1}{106}\right)\)
= \(3.\frac{101}{530}\)
= \(\frac{303}{530}\)
Đặt \(A=\frac{11}{2}.\frac{12}{2}.\frac{13}{2}...\frac{20}{2}\)
\(=\frac{\left(11.13.15.17.19\right).12.14.16.18.20}{2^{10}}\)
\(\frac{\left(11.13.15.17.19\right).\left(3.2^2\right).\left(7.2^1\right).2^4.\left(9.2^1\right).\left(5.2^2\right)}{2^{10}}\)
\(=\frac{\left(1.3.5.7.9.11.13.15.17.19\right).2^{10}}{2^{10}}\)
a)đặt B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
b,c tự làm
Ta có: \(A=\frac{2}{11}+\frac{2}{12}+\frac{2}{13}+\cdots+\frac{2}{40}\)
Mà \(\frac{2}{11}+\frac{2}{12}+\frac{2}{13}+\ldots+\frac{2}{20}>\frac{2}{20}+\frac{2}{20}+\cdots+\frac{2}{20}=10\times\frac{2}{20}=1\)
Mà \(\) \(\frac{2}{21}+\frac{2}{22}+\frac{2}{23}+\ldots+\frac{2}{30}>\frac{2}{30}+\frac{2}{30}+\cdots+\frac{2}{30}=10\times\frac{2}{30}=\frac23\)
Mà \(\frac{2}{31}+\frac{2}{32}+\frac{2}{33}+\ldots+\frac{2}{40}>\frac{2}{40}+\frac{2}{40}+\cdots+\frac{2}{40}=10\times\frac{2}{40}=\frac12\)
Cộng cả ba vế ta đươc: \(A>1+\frac23+\frac12=\frac{13}{6}\) (đpcm)