356,5 + 125,3 + 450,1 

= 481,8 + 450,1

= 931,9

Vậy 356,5 + 125,3 + 450,1 = 1031,9 là sai

   \(\dfrac{3^{11}.11+3^{11}.21}{3^9.2^5}\)

=  \(\dfrac{3^{11}.\left(11+21\right)}{3^932}\)

= \(\dfrac{3^{11}32}{3^9.32}\)

= 3²

= 9 

Khi em hỏi câu này thì chính em cũng đang sử dụng thông tin internet đó em.

2 - \(x\) = 17 - (-5)

2 - \(x\) = 17 + 5

2 - \(x\) = 22

     \(x\) = 2 - 22

      \(x\) = -20

      B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)

3 \(\times\)B =  1 + \(\dfrac{1}{3}\) +   \(\dfrac{1}{9}\) +  \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)

3 \(\times\) B - B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) -(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+...+\(\dfrac{1}{729}\))

2B          = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{9}\)-\(\dfrac{1}{27}\)-...- \(\dfrac{1}{729}\)

2B          = 1 - \(\dfrac{1}{729}\)

2B         = \(\dfrac{728}{729}\)

   B = \(\dfrac{728}{729}\) : 2

   B  = \(\dfrac{364}{729}\)

           A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)

     2 \(\times\) A = 1   + \(\dfrac{1}{2}\) +  \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)

 2 \(\times\) A - A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\))

        A      = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\)

        A       =  1 - \(\dfrac{1}{32}\)

        A       =   \(\dfrac{31}{32}\)

Thiếu vế phải rồi em nhé!

   (-23).29 - 29.56 + 29.(-21)

= - 29.( 23 + 56 + 21)

=  -29 . 100

=    - 2900