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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) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)
\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)
H
0
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
VN
3
16 tháng 8 2020
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) \(25^{25}+5^{49}-125^{16}=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}=5^{48}\left(5^2+5-1\right)\)
\(=5^{48}\cdot29⋮29\left(đpcm\right)\)
NT
1
13 tháng 9 2020
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)
c) Đề hơi sai roi bạn oi
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)
28 tháng 11 2017
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)
\(A=\frac{7^{49}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
\(A=\frac{7^{48}.5^{30}.2^8\left(1-28\right)}{5^{29}.2^8.7^{48}}\)
\(A=5.\left(-27\right)\)
\(A=-135\)
\(\frac{25^{25}.7^{30}}{5^{48}.49^{16}}=\frac{\left(5^2\right)^{25}.7^{30}}{5^{48}.\left(7^2\right)^{16}}=\frac{5^{50}.7^{30}}{5^{48}.7^{32}}=\frac{5^2}{7^2}=\frac{25}{49}\)
hok tốt!!!
\(\frac{25^{25}.7^{30}}{5^{48}.49^{16}}=\frac{\left(5^2\right)^{25}.7^{30}}{5^{48}.\left(7^2\right)^{16}}=\frac{5^{30}.7^{30}}{5^{48}.7^{32}}=\frac{5^{48}.5^2.7^{30}}{5^{48}.7^{30}.7^2}=\frac{25}{49}\)