Ta đặt A=\(-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(\Rightarrow A=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)= \(-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\) 

= - \(\left(1-\dfrac{1}{10}\right)=-\left(\dfrac{10-1}{10}\right)=-\dfrac{9}{10}\)

Ta có: \(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=-\left(-\dfrac{1}{10}+1\right)\)

\(=-\left(1-\dfrac{1}{10}\right)\)

\(=-\left(\dfrac{10}{10}-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)

= - ( 1/2 +1/6+1/12+1/20+ 1/30+ 1/42+ 1/56+ 1/72+ 1/90)

= - ( 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)

= - ( 1- 1/10 )

= -9/10

\(-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\left(-\frac{1}{90}\right)+\left(-\frac{1}{72}\right)+\left(-\frac{1}{56}\right)+......+\left(-\frac{1}{2}\right)\)

\(=\left(-1\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{9.10}\right)\)

\(=\left(-1\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\left(-1\right).\left(1-\frac{1}{90}\right)=\frac{\left(-1\right).89}{90}=-\frac{89}{90}\)

nha

a= -8 phần 9

b= 2

a.A=-1/2-1/6-1/12-1/20-1/30-1/42-1/56-1/72=-151/180

Vậy A=151/180

b.B=0,5+0,4+1/3+1/6+5/7-4/35=2

Vậy B=2

B= \(-\frac{9}{10}=-0,9\)

\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)

\(\frac{-1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

= \(-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

=\(-\left(1-\frac{1}{10}\right)=-\left(\frac{9}{10}\right)=-\frac{9}{10}\)

\(\frac{-1}{90}-\frac{-1}{72}-\frac{-1}{56}-\frac{-1}{42}-\frac{-1}{30}-\frac{-1}{20}-\frac{-1}{12}-\frac{-1}{6}-\frac{-1}{2}\)

\(=\frac{-1}{10.9}-\frac{-1}{9.8}-\frac{-1}{8.7}-\frac{-1}{7.6}-\frac{-1}{6.5}-\frac{-1}{5.4}-\frac{-1}{4.3}-\frac{-1}{3.2}-\frac{-1}{2.1}\)

\(=\frac{79}{90}\)