\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)

\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)

\(D=1+-1+\dfrac{1}{41}\)

\(D=0+\dfrac{1}{41}\)

\(D=\dfrac{1}{41}\)

\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)

=1/57

\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)

\(-\dfrac{1}{2}-\dfrac{2}{5}-\dfrac{-1}{2}.\dfrac{5}{7}-\dfrac{1}{6}+\dfrac{4}{35}-\dfrac{1}{41}\)

=\(-\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{5}{14}-\dfrac{1}{6}+\dfrac{4}{35}-\dfrac{1}{41}=-\dfrac{1067}{1722}\)

có ngoặc ko

Các câu dễ tự làm nha:

\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)

a) \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{12}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{4}+\dfrac{-2}{5}=\dfrac{-3}{20}\)

b) \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}=\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+\left(\dfrac{-1}{5}-\dfrac{-7}{10}\right)+\dfrac{3}{4}\)

\(=\dfrac{-3}{2}+\dfrac{1}{2}-\dfrac{3}{4}\)

= \(=-1-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

c)\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)

= \(\left(\dfrac{1}{2}-\dfrac{-1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-4}{35}+\dfrac{5}{7}-\dfrac{-2}{5}\right)+\dfrac{1}{41}\)

= \(1+1+\dfrac{1}{41}\)

= \(\dfrac{83}{41}\)

d)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

= \(\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}\)

= \(\dfrac{1}{100}-\dfrac{1}{1}\)

= \(\dfrac{-99}{100}\)

d đảo 1/1.2.1/2.3 ... 1/99.1000

=1/1 -1/2 +1/2-1/3 ... -1/99 - 1/1000

=1/1 -1/1000

=999/1000

có j sai thì thông cảm nhé

1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

\(A=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{5}{7}+\dfrac{1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\)

\(=\dfrac{3+2+1}{6}+\dfrac{14+25-4}{35}+\dfrac{1}{41}\)

\(=1+\dfrac{1}{41}+1=2+\dfrac{1}{41}=\dfrac{83}{41}\)

A=\(\dfrac{1}{2}\)-\(\left(\dfrac{-2}{5}\right)\)+\(\dfrac{1}{3}\)+\(\dfrac{5}{7}\)-\(\left(\dfrac{-1}{6}\right)\)+\(\left(\dfrac{-4}{35}\right)\)+\(\dfrac{1}{41}\)

=\(\dfrac{1}{2}\)+\(\dfrac{2}{5}\)+\(\dfrac{1}{3}\)+\(\dfrac{5}{7}\)+\(\dfrac{1}{6}\)-\(\dfrac{4}{35}\)+\(\dfrac{1}{41}\)

=\(\left[\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right]\)+\(\left[\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right]\)+\(\dfrac{1}{41}\)

= 1 + 1 +\(\dfrac{1}{41}\)

= \(\dfrac{83}{41}\)

\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)