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a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)
\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)
\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)
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Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
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a)\(12^3.3^3\)
\(=\left(12.3\right)^3\)
\(=36^3\)
\(=46656\)
b)\(2.8^4\)
\(=2.\left(2^3\right)^3\)
\(=2.2^9\)
\(=2^{10}\)
\(=1024\)
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1+1+1+1+1+1+1+1+1+1=10
1+1+1+1+1+1+1+1+1-1=9
1+1+1+1+1+1+1+1-1-1=8
1+1+1+1+1+1+1-1-1-1=7
1+1+1+1+1+1-1-1-1-1=6
1+1+1+1+1-1-1-1-1-1=5
1+1+1+1-1-1-1-1-1-1=4
1+1+1-1-1-1-1-1-1-1=3
1+1-1-1-1-1-1-1-1-1=2
1-1-1-1-1-1-1-1-1-1-1=0
Nối cung ta điền dấu cộng hết
Các số còn lại ta giảm mỗi số một dấu cộng = dấu -
100km
cần lm hộ ko
mà rem và rem sai xùi