Tính 1 cách hợp lí :
1+ 1/5 + 1/25 + 1/125 + 1/625 +........+ 1/78125
1,thực hiện phép tính bằng cách thuận tiện nhất
a,1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)
\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)
\(1+\frac{1}{5}+\frac{1}{25}+\frac{1}{125}+\frac{1}{625}+.......+\frac{1}{78125}\)
tính hộ nha
1/tính hợp lý ( nếu có thể)
a. 125 *37+125*64-78125:625
=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) A=1+1/5+1/25+1/125+1/625+...+1/78125
b)B=1/3+1/12+1/48+1/192+1/768+...+1/36864
c)C=5/3+5/6+5/12+5/24...+5/192+5/384
GIÚP MÌNH VỚI XIN MỌI NGƯỜI AI GIÚP MÌNH THÌ MÌNH CHO TICK LÀM TẤT CẢ MỌI THỨ
1. tính bằng cách hợp lí :
a. 5 . 125 . 2 . 41 . 8 = c. 8 . 12 . 125 . 2 =
b. 25 . 7 . 10 . 4 = d. 4 . 36. 25 . 50 =
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!! ^^
Thực hiện chiến phép tính
E=1/9-1/7-1/11/4/9-4/7-4/11+3/5-3/25-3/125-3/625/4/5-4/25-4/125-4/625
S=1+1/5+1/25+1/125+1/625
\(S=\dfrac{625}{625}+\dfrac{125}{625}+\dfrac{25}{625}+\dfrac{5}{625}+\dfrac{1}{625}\)
\(=\dfrac{781}{625}\)
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) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)
2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)
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:
\(S=\left(\dfrac{1}{5}\right)^0+\left(\dfrac{1}{5}\right)^1+...+\left(\dfrac{1}{5}\right)^7\)
\(u1=1;q=\dfrac{1}{5}\)
\(S_7=\dfrac{1\cdot\left(1-q^7\right)}{1-q}=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}=\dfrac{5}{4}\left(1-\dfrac{1}{5^7}\right)\)
1) tính \(S=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{19683}\)
2) tính \(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{78125}\)
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
\(S_7=\dfrac{1-\left(\dfrac{1}{5}\right)^7}{1-\dfrac{1}{5}}\approx1,25\)