4^3.3^4.9^2
6^12
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
RM
1
26 tháng 7 2016
a, \(4^7.3^4.9^6:6^{13}\)
\(=\left(2^{14}.3^4.3^{12}\right):\left(2^{13}.3^{13}\right)\)
\(=2^{14}:2^{13}.3^{16}:3^{13}\)
\(=2.3^3=54\)
b, \(2^3.3^2-5^{16}:25^7\)
\(=72-5^{16}:5^{14}\)
\(=72-5^2=47\)
LH
1
26 tháng 7 2016
4^7.3^4.9^6:6^13=4^7.3^4.(3^2)^6:6^13
=4^7.3^16:3^13.2^13
=(2^2)^7.3^16:3^13.2^13
=2^14.3^16:3^13.2^13
=2.3^3
=54
28 tháng 6 2021
Ta có : \(C=\dfrac{4^6.3^4.9^5}{6^{12}}=\dfrac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}=\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=3^{4+10-12}=3^2=9\)
21 tháng 9 2016
\(\left(4096.81.729\right):2176782336\)
\(\left(331776.729\right):2176782336\)
\(\left(241864704\right):2176782336\)
21 tháng 9 2016
\(\frac{4^6.3^4.9^3}{6^{12}}=\frac{2^{12}.3^4.3^6}{2^{12}.3^{12}}=\frac{2^{12}.3^{10}}{2^{12}.3^{12}}=\frac{1}{3^2}=\frac{1}{9}\)
DT
3
29 tháng 7 2017
B = 4^9.36+64^4
16^4.100
B = 2^20.(9+16)
2^18.5^2
B = 2^20.5^2
2^18.5^2
B = 2^2
B = 4
D = 4^6.3^4.9^5
6^12
D = 2^12.3^4.9^5
2^12.3^12
D = 2^12.3^14
2^12.3^12
D = 3^2
D = 9
29 tháng 7 2017
\(\frac{4^9.36+64^4}{16^4.100}=\frac{4^{10}\left(9+8\right)}{4^{10}.25}=\frac{17}{25}\)
DT
0
18 tháng 9 2021
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
CC
0