\(A=19^{5^{1^{8^{9^0}}}}\)\(+2^{9^{1^{9^{6^9}}}}\)
\(=19^{5^1}+2^{9^1}\)
\(=19^5+2^9\)
\(=...9+512\)
\(=...1\)
Vậy chữ số tận cùng của A là 1
Tính giá trị của biểu thức(tính nhanh nếu có thể):
1)\(\frac{35.\left(27^{8^{ }}+2.9^{11}\right)}{15.\left(81^{6^{ }}-12.3^{19}\right)}\)
2)\(\frac{5.4^{15^{ }}.9^{9^{ }}-4.3^{20^{ }}.8^9}{^{ }5.2^{9^{ }}.6^{19^{ }}-7.2^{29^{ }}.27^6}\)
3)\(\frac{4^{6^{ }}.9^{5^{ }}+6^{9^{ }}.120}{-8^{4^{ }}.3^{12^{ }}-6^{11}}\)
tính
1, 125^5:25^7
2, 27^8:9^9
3,36^5:6^8
4, 49^6:7^10
5, 7^20:49^9
6, 1/2^10:1/8^3
7, (-1/2)^21:1/4^10
8,1/16^5:(-1/2)^18
9,1/5^30:1/25^14
10, 1/125^6:1/25^14
4^5 . 9^4 - 2 . 6^9 / 2^10 . 3^8 + 6^8 . 20
C/m rằng : a,( 10^19+10^18+10^17 ) chia hết 555; b, ( 81 ^7 - 27^9 - 9^13) chia hết cho 15; c, ( 4^13 + 32^5 - 8^8 ) chia hết cho 5 cảm ơn mn
a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)
Bài 1:
\(\frac{8^{18}+1}{8^{19}+1}\)
Bài 2:
a) Chứng minh: \(9^{1945}-2^{1930}\) chia hết cho 5
b) Chứng minh: \(4^{2010}+2^{2014}\) chia hết cho 10
chứng minh rằng
3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+.....+19/9^2.10^2<1
Tính
A=\(\dfrac{2^{19}\cdot27^3-15\cdot\left(-4\right)^9\cdot9^4}{6^9\cdot2^{10}+\left(-12\right)^{10}}\)
Tính
\(\dfrac{-15}{8}-\dfrac{4}{9}+\dfrac{-1}{8}+\dfrac{-5}{7}+\dfrac{-5}{9}\)
\(\left(\dfrac{-3}{4}\right)^3-\left(\dfrac{1}{2}\right)^6+2\)
-6,25+ 2,475- 12,123
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