Ta có:

\(\begin{array}{l}\frac{8}{{32}} = \frac{{8:8}}{{32:8}} = \frac{1}{4};\\\frac{{13}}{{54}};\\\frac{{ - 9}}{{ - 36}} = \frac{{( - 9):( - 9)}}{{( - 36):( - 9)}} = \frac{1}{4}\end{array}\)

Như vậy, ta có dãy tỉ số bằng nhau là: \(\frac{1}{4} = \frac{8}{{32}} = \frac{{ - 9}}{{ - 36}}\).

\(\dfrac{1}{4}=\dfrac{8}{32}=\dfrac{-9}{-36}\)

\(\frac{21}{36}x\frac{39}{14}x\frac{54}{13}=\frac{21x39x54}{36x14x13}=\frac{3x7x13x3x9x6}{6x6x7x2x13}=\frac{3x9}{2x2}=\frac{27}{4}\)

\(\frac{21}{36}x\frac{39}{14}x\frac{54}{13}=\frac{13}{8}x\frac{54}{13}=\frac{54}{8}=\frac{27}{4}\)

Vì bớt cả trên lẫn dưới đi 13

Bằng 27 phần 4

\(\frac{12}{67}\)+ \(\frac{34}{54}\)= \(\frac{1463}{1809}\)

Nhớ k mk đấy !!!!!!!!!!!

12/67 + 34/54 = 12/67 + 12/27 = 12x27/67x27 + 34x67/27x67 = 324/1809 + 2278/1809 = 324 + 2278/1809 = 2602/1809

vậy 12/67 + 34/54 = 2602/1809

\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)

\(=\frac{1}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)

\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(=\frac{1}{3}.\frac{10}{33}\)

\(=\frac{10}{99}\)

\(\frac{1}{18}+\frac{1}{54}+...+\frac{1}{990}\)

\(=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)

\(=\frac{1}{3}\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{10}{99}\)

A = 1/20

B = 85/504

               k cho mình nha 

giải thích giúp mình đi Bui Tra My

Ta có: \(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)

\(A=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)-\frac{3}{4}-\frac{1}{36}+\frac{1}{73}\)

\(A=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\frac{1}{73}\)

\(A=\frac{1}{9}+\frac{10}{15}-\frac{7}{9}+\frac{1}{73}\)

\(A=\frac{1}{9}+\frac{2}{3}-\frac{7}{9}+\frac{1}{73}\)

\(A=\frac{1}{9}+\frac{6}{9}-\frac{7}{9}+\frac{1}{73}\)

\(A=\frac{7}{9}-\frac{7}{9}+\frac{1}{73}\)

\(A=\frac{1}{73}\)

Vậy: \(A=\frac{1}{73}\)

cho mik hỏi k kiểu j z

a) 54x113+45x113+113                                          b)117x(36+62) -17x(62+36)
   =(54+45+1)x113                                                    =117x98-17x98
   = 100x113                                                            =(117-17)x98
   = 11300                                                                =100x98
                                                                               =9800

a,54 x 113 + 45 x 113 + 113

=(54 + 45 +1) x 113

=100 x 113 = 11300

a) 113*( 54+45+1 ) = 113* 100=11300

b) 117*98-17*98=98*( 117-17 ) = 98*100=9800

`1, (10-4-6)+(5/4-9/14-5/7+7/3)

=0+187/84

=187/ 84

29/32.41/36-29/32.32/58-41/36.29/32-41/36.18/41

=-29/32.32/58-41/36.18/41

=-1/2-1/2=-1