\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)

\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)

\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)

\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)

\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)

thích xem thiếu nữ toàn phong lắm hả

=125.(37+64)-125

=125.101-125

=125.100+125.1-125

=12500+125-125

=12500+0

=12500

125 . 37 + 125 . 64 - 78125 : 625

= 125 . 37 + 125 . 64 - 125

= 125 . (37 + 64 - 1)

= 125 . 100

= 12500

bạn chú ý dến câu trả lời của mk là dấu . là dấu nhân

a) 5 . 125 . 2 . 41 . 8

= (5 . 8) . (125 . 2) . 41

= 40 . 250 . 41

= 10 000 . 41

= 410 000

b) 25 . 7 . 10 . 4

= (25 . 4 ) . (10 . 7)

= 100 . 70

= 7000

Chúc bạn học tốt!! ^^

a,5.125.2.41.8=(125.8).(2.5).41=1000.10.41=410000

b,25.7.10.4=(25.4).70=100.70=7000

c,8.12.125.2=(8.125).12.2=100.24=2400

d,4.36.25.50=(4.25).(2.50).18=100.100.18=180000

c) 8 . 12 . 125

= (125 . 8) . 12

= 1000 . 12

= 12 000

d) 4 . 36 . 25 . 50

= (25 . 4) . (50 . 36)

= 100 . 1800

= 180 000

Chúc bạn học tốt!! ^^

\(S=\dfrac{625}{625}+\dfrac{125}{625}+\dfrac{25}{625}+\dfrac{5}{625}+\dfrac{1}{625}\)

\(=\dfrac{781}{625}\)

   S       =            1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\) + \(\dfrac{1}{625}\)

5.S        =       5 +1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\)

5S  - S  =        5 - \(\dfrac{1}{625}\)

   S       =        ( 5 - \(\dfrac{1}{625}\)) : 4

   S =      \(\dfrac{781}{625}\) 

1, Ta có \(\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3};\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3};...\)

-> Là cấp số nhân, q = 1/3 

Ta có \(S_9=1.\dfrac{1-\left(\dfrac{1}{3}\right)^9}{1-\left(\dfrac{1}{3}\right)}\approx1,5\)

b, Ta có \(\dfrac{\dfrac{1}{5}}{1}=\dfrac{1}{5};\dfrac{\dfrac{1}{25}}{\dfrac{1}{5}}=\dfrac{1}{5};...\)

-> Là cấp số nhân, q = 1/5 

\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)

1: \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^9}\)

\(=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^9\)

u1=1; q=1/3

\(S_9=\dfrac{u1\cdot\left(1-q^9\right)}{1-q}=\dfrac{1\left(1-\left(\dfrac{1}{3}\right)^9\right)}{1-\dfrac{1}{3}}\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3^9}\right)\)

2:

\(S=\left(\dfrac{1}{5}\right)^0+\left(\dfrac{1}{5}\right)^1+...+\left(\dfrac{1}{5}\right)^7\)

\(u1=1;q=\dfrac{1}{5}\)

\(S_7=\dfrac{1\cdot\left(1-q^7\right)}{1-q}=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}=\dfrac{5}{4}\left(1-\dfrac{1}{5^7}\right)\)