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\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow A=\frac{2}{5}\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\frac{9}{10}=0\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(\Leftrightarrow A=\frac{9}{10}-\frac{9}{10}\)
\(\Leftrightarrow A=0\)
a) 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/10.9 - 1/9.8 - 1/8.7 - 1/7.6 - 1/6.5 - 1/5.4 - 1/4.3 - 1/3.2 - 1/2.1
= 1/10 - 1
= 0,1 - 1
= -0,9
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
=\(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{9}{10}-(\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1})\)
=\(\frac{9}{10}-(\frac{1}{10}+\frac{1}{9}-\frac{1}{9}+\frac{1}{8}-\frac{1}{8}+\frac{1}{7}-\frac{1}{7}+\frac{1}{6}-\frac{1}{6}+\frac{1}{5}-\frac{1}{5}+\frac{1}{4}-\frac{1}{4}+\frac{1}{3}-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+\frac{1}{1})\)
= \(\frac{9}{10}-\left(\frac{1}{10}+\frac{1}{1}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{10}+\frac{10}{10}\right)\)
=\(\frac{9}{10}-\frac{11}{10}\)
= \(\frac{-2}{10}=-\frac{1}{5}\)
# Chúc bạn học tốt #
Ta có : \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{9.10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\frac{9}{10}=0\)
Có \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{90}=\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{9.10}\)
= \(1-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{9}-\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
kết quả là \(\frac{1}{10}\)nhà mình không biết cách làm vậy cho mình xin lỗi nha!
mình trả lời đầu tiên đó nha!