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\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)
= 8
A= 1/10+1/11+1/12+1/13+...........+1/99+1/100
2A=1/9+1/10+1/11+1/12+...........+1/98+1/99
2A-A=(1/10+1/11+1/12+1/13+.............+1/99+1/100)-(1/9+1/10+1/11+1/12+............1/98+1/99)
A=1/100-1/9
a)
\(\dfrac{2}{3}x-\dfrac{1}{5}x=-\dfrac{7}{10}\\ \left(\dfrac{2}{3}-\dfrac{1}{5}\right)x=-\dfrac{7}{10}\\ \dfrac{7}{15}x=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}:\dfrac{7}{15}\\ x=-\dfrac{15}{10}\\ x=-\dfrac{3}{2}.\)
b)
\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right).\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2021}-1\right)\\ =\dfrac{-9}{10}.\dfrac{-10}{11}.\dfrac{-11}{12}...\dfrac{-2020}{2021}\\ =\dfrac{9}{2021}.\)
c)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ =\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{3}-\dfrac{1}{10}\\ =\dfrac{7}{30}.\)
(câu c em ghi nhầm đề 1 chút nhé, chỗ đó là 56 chứ không phải 50)
99/100