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\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)
\(D=12\cdot\frac{49}{303}\)
\(D=\frac{588}{303}\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
-5/ 12. (2/11+ 9/11+ (-1)
=-5/ 12. [(11/11+(-1)]
=-5/ 12. (1-1)
= -5/12. 0
=0
\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)
\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=\dfrac{1458}{5}.\dfrac{2}{9}\)
\(=\dfrac{324}{5}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)
\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)
\(\dfrac{12}{6.7}+\dfrac{12}{7.22}+\dfrac{12}{22.15}+\dfrac{12}{15.38}+...+\dfrac{12}{97.202}\)
\(=12.\left(\dfrac{1}{6.7}+\dfrac{1}{7.22}+\dfrac{1}{22.15}+\dfrac{1}{15.38}...+\dfrac{1}{97.202}\right)\)
\(=12.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{15}+...+\dfrac{1}{97}-\dfrac{1}{202}\right)\)
\(=12.\left(\dfrac{1}{6}-\dfrac{1}{202}\right)\)
\(=\dfrac{196}{101}\)