gọi x là số cần tìm

 X:21=10(dư14)

X= 10.21+14

X= 210+14

X= 224

Vậy số cần tìm là: 224

giúp mik với

2010/2015 nhé

\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}\)\(+\dfrac{89}{90}\)

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

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

=9-\(\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\right)\)

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

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

=9-\(\dfrac{9}{10}\)

=\(\dfrac{81}{10}\)

vì phần cuối bảng có hạn nên đã mất dấu ngoặc bạn thông cảm

a.E={12;15;18;21}

b.E={2010;2013;2019}

c.E={1235}

=> (x-11):3 - 273=91 => (x-11):3=364 => x-11=1092 => x=1103

(x-11):3-21.13=91

(x-11):3-273=91

(x-11):(-270)=91

(x-11)           =91.(-270)

(x-11)           =(-24570)

x                  =(-24570)+11

x                  =-24550