Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$

$=2.3+2^3.3+...+2^{99}.3$

$=3(2+2^3+...+2^{99})\vdots 3$

Ta có đpcm.

a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)

= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)

= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)

= \(\dfrac{4}{21}\)

b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)

   =  - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)

   = -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)

  = \(\dfrac{-61}{30}\) 

c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)

 =  - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)

= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)

=  \(\dfrac{2}{21}\)

d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)

 = - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)

 = - \(\dfrac{21}{77}\) -  \(\dfrac{22}{77}\)

= - \(\dfrac{43}{77}\)

a, Ta có :

 A =  1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100

2A =  2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101

A = 2A – A =  ( 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 ) –( 1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 )

=  2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 – 1 - 2 - 2 2 - 2 3 - 2 4 - . . . - 2 99 - 2 100

=  2 101 - 1

Vậy A =  2 101 - 1

b, Ta có.

B = 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99

5 2 B =  5 2 ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )

25B =  5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101

25B – B = ( 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 ) –  ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )

24B =  5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 – 5 - 5 3 - 5 5 - . . . - 5 97 - 5 99

24B =  5 101 - 5

B =  5 101 - 5 24 = 5 5 100 - 1 24

Vậy B =  5 5 100 - 1 24

Ta có: \(A=\dfrac{-5}{7}:\left(25-24\dfrac{4}{7}\right)+15\dfrac{1}{3}\cdot\dfrac{-1}{23}\)

\(=\dfrac{-5}{7}:\left(25-\dfrac{172}{7}\right)+\dfrac{46}{5}\cdot\dfrac{-1}{23}\)

\(=\dfrac{-5}{7}\cdot\dfrac{7}{3}+\dfrac{-2}{5}\)

\(=\dfrac{-5}{3}+\dfrac{-2}{5}\)

\(=\dfrac{-31}{15}\)

mik đang cần gấp

giúp mik với ☹

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)

\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{23.24.25}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{23.24}-\frac{1}{24.25}\)

\(=\frac{1}{1.2}-\left(\frac{1}{2.3}-\frac{1}{2.3}\right)-\left(\frac{1}{3.4}-\frac{1}{3.4}\right)-\left(\frac{1}{4.5}-\frac{1}{4.5}\right)-....-\left(\frac{1}{23.24}-\frac{1}{23.24}\right)-\frac{1}{24.25}\)

\(=\frac{1}{1.2}-\frac{1}{24.25}\)

\(=\frac{1}{2}-\frac{1}{600}=\frac{299}{600}\)

Vậy : \(A=\frac{299}{600}\)