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\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)
làm được câu a thui
\(\frac{1}{100}=\frac{1}{10.10}\);\(\frac{1}{90}=\frac{1}{9.10}\);...
Suy ra \(\frac{1}{10.10}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-...-\frac{1}{1.2}\)
\(\frac{1}{10}-\frac{1}{10}-\frac{1}{10}-\frac{1}{9}-...-1-\frac{1}{2}\)
\(\frac{1}{10}-\frac{1}{2}\)
\(-\frac{4}{10}\)
= 1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
= 1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
= 9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
= 9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
= 9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
= 9 – (1 – 1/10) = 9 – 9/10 = 81/10
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
\(=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{9.10}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
9 số 1
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}\)
\(=\frac{81}{10}\)
Ủng hộ mk nha ^-^
\(\frac{-1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-\left(1-\frac{1}{10}\right)=-\left(\frac{9}{10}\right)=-\frac{9}{10}\)
\(\frac{-1}{90}-\frac{-1}{72}-\frac{-1}{56}-\frac{-1}{42}-\frac{-1}{30}-\frac{-1}{20}-\frac{-1}{12}-\frac{-1}{6}-\frac{-1}{2}\)
\(=\frac{-1}{10.9}-\frac{-1}{9.8}-\frac{-1}{8.7}-\frac{-1}{7.6}-\frac{-1}{6.5}-\frac{-1}{5.4}-\frac{-1}{4.3}-\frac{-1}{3.2}-\frac{-1}{2.1}\)
1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833