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Ta có: \(19\equiv-1\left(mod5\right)\)
\(\Rightarrow\hept{\begin{cases}19^{45}\equiv-1\left(mod5\right)\\19^{30}\equiv1\left(mod5\right)\end{cases}}\)
\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod5\right)\)
\(\Rightarrow19^{45}+19^{30}⋮5\) \(\left(1\right)\)
Lại có: \(19\equiv-1\left(mod4\right)\)
\(\Rightarrow\hept{\begin{cases}19^{45}\equiv-1\left(mod4\right)\\19^{30}\equiv1\left(mod4\right)\end{cases}}\)
\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod4\right)\)
\(\Rightarrow19^{45}+19^{30}⋮4\) \(\left(2\right)\)
Mà \(\left(4;5\right)=1\) \(\left(3\right)\)
Từ (1);(2) và (3) suy ra
\(19^{45}+19^{30}⋮4.5=20\)
Ta có: 19≡−1(mod5)
⇒{
1945≡−1(mod5) |
1930≡1(mod5) |
⇒1945+1930≡0(mod5)
⇒1945+1930⋮5 (1)
Lại có: 19≡−1(mod4)
⇒{
1945≡−1(mod4) |
1930≡1(mod4) |
⇒1945+1930≡0(mod4)
⇒1945+1930⋮4 (2)
Mà (4;5)=1 (3)
Từ (1);(2) và (3) suy ra
1945+1930⋮4.5=20
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
\(A,\frac{4^9.36+64}{16^4.100}=\frac{\left(2^2\right)^9.2^2.3^2+2^6}{\left(2^4\right)^4.2^2.5^2}=\frac{2^{20}.3^2+2^6}{2^{18}.5^2}=\frac{2^6\left(2^{14}.3^2+1\right)}{2^{18}.5^2}=\frac{2^{14}.3^2+1}{2^{12}.5^2}=\frac{147457}{102400}\)
B,
\(\frac{11.3^{22}.3-9^{13}}{\left(2.3^{14}\right)^2}=\frac{11.3^{22}-\left(3^2\right)^{13}}{2^2.3^{28}}=\frac{11.3^{22}-3^{26}}{2^2.3^{28}}=\frac{3^{22}\left(11.1-3^4\right)}{2^2.3^{28}}=\frac{11-81}{2^2.3^6}=-\frac{70}{2916}=-\frac{35}{1456}\)
c,
\(\frac{45^3.20^4.18}{180^5}=\frac{\left(3^2.5\right)^3.\left(5.2^2\right)^4.2.3^2}{\left(2^2.3^2.5\right)^5}=\frac{3^6.5^3.5^4.2^8.2.3^2}{2^{10}.3^{10}.5^5}=\frac{3^8.2^{10}.5^7}{2^{10}.3^{10}.5^5}=\frac{5^2}{3^2}=\frac{25}{9}\)
\(\frac{4^9\cdot36+64}{16^4\cdot100}=\frac{2^6\cdot147457}{2^{16}\cdot100}=\frac{147457}{2^{10}\cdot100}\)
\(\frac{11\cdot3^{22}\cdot3-9^{13}}{2^2\cdot3^{28}}=\frac{3^{23}\left(11-3^3\right)}{2^2\cdot3^{28}}=\frac{-16\cdot3^{23}}{2^2\cdot3^{28}}=\frac{-4}{243}\)
\(\frac{45^3\cdot20^4\cdot18}{180^5}=\frac{3^8\cdot2^9\cdot5^7}{2^{10}\cdot3^{10}\cdot5^5}=\frac{25}{18}\)
220-45 = 2-25