VN
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TQ
3
24 tháng 9 2020
\(9^7+81^4-27^5\)
\(=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}.\left(1+3^2-3\right)\)
\(=3^{14}.7⋮7\)
=> đpcm
\(25^{25}+5^{49}-125^{16}\)
\(=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}\)
\(=5^{48}.\left(5^2+5-1\right)\)
\(=5^{48}.29⋮29\)
=> đpcm
B
24 tháng 9 2020
Bài làm :
\(\text{1) }9^7+81^4-27^5\)
\(=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}\left(1+3^2-3\right)\)
\(=3^{14}.7⋮7\)
=> Điều phải chứng minh
\(\text{2)}25^{25}+5^{49}-125^{16}\)
\(=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}\)
\(=5^{48}\left(5^2+5-1\right)\)
\(=5^{48}.29⋮29\)
=> Điều phải chứng minh
TG
3 tháng 8 2020
1) \(125^5:25^7\)
\(=\left(5^3\right)^5:\left(5^2\right)^7\)
\(=5^{15}:5^{14}\)
= 5
2) \(27^8:9^9\)
\(=\left(3^3\right)^8:\left(3^2\right)^9\)
\(=3^{24}:3^{18}\)
\(=3^6\)
3) \(36^5:6^8\)
\(=\left(6^2\right)^5:6^8\)
\(=6^{10}:6^8\)
\(=6^2\)
4) \(49^6:7^{10}\)
\(=\left(7^2\right)^6:7^{10}\)
\(=7^{12}:7^{10}=7^2\)
5) \(7^{20}:49^9\)
\(=7^{20}:\left(7^2\right)^9\)
\(=7^{20}:7^{18}=7^2\)
6) \(\frac{1}{2^{10}}:\frac{1}{8^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{\left(2^3\right)^3}\)
\(=\frac{1}{2^{10}}:\frac{1}{2^9}=\frac{1}{2^{10}}.\frac{2^9}{1}=\frac{1}{2}\)
7) \(\left(-\frac{1}{2}\right)^{21}:\frac{1}{4^{10}}\)
\(=\frac{\left(-1\right)^{21}}{2^{21}}:\frac{1}{\left(2^2\right)^{10}}\)
\(=-\frac{1}{2^{21}}:\frac{1}{2^{20}}=-\frac{1}{2^{21}}.\frac{2^{20}}{1}\)
\(=-\frac{1}{2}\)
8) \(\frac{1}{16^5}:\left(-\frac{1}{2}\right)^{18}\)
\(=\frac{1}{\left(2^4\right)^5}:\frac{\left(-1\right)^{18}}{2^{18}}\)
\(=\frac{1}{2^{20}}:\frac{1}{2^{18}}\)
\(=\frac{1}{2^{20}}.\frac{2^{18}}{1}=\frac{1}{4}\)
9) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)
T
10 tháng 9 2023
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
TL
B={[5^10. 7^3 - 25^2 . 49^2] : [(125 . 7)^3+5^9 . 14^3]} - {[2^12-4^6 . 9^2] : [(2^2.3)^6+8^4 . 3^5]
1
9 tháng 7 2023
Sửa đề: \(5^9\cdot49^2\)
\(=\dfrac{5^{10}\cdot7^3-5^9\cdot7^4}{5^9\cdot7^3+5^9\cdot14^3}-\dfrac{2^{12}-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\dfrac{5^9\cdot7^3\left(5-7\right)}{5^9\cdot7^3\left(1+8\right)}-\dfrac{2^{12}\left(1-3^4\right)}{2^{12}\left(3^6+3^5\right)}=\dfrac{-2}{9}+\dfrac{80}{972}\)
=-34/243
Mình đang cần đáp án gấp mọi người giúp mình nha
a) \(9^7+81^4-27^5=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}\left(1+3^2-3\right)=3^{14}\cdot7⋮7\left(đpcm\right)\)
b) \(25^{25}+5^{49}-125^{16}=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}=5^{48}\left(5^2+5-1\right)\)
\(=5^{48}\cdot29⋮29\left(đpcm\right)\)