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\(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)
\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)
\(=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}\)
\(=\dfrac{5\cdot\left(-3\right)}{4\cdot6}=\dfrac{-15}{24}=\dfrac{-5}{8}\)
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\(Q=\dfrac{8^5.\left(-5\right)^8-\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(Q=\dfrac{2^{15}.5^8+2^5.2^9.5^9}{2^{16}.5^7+4^8.5^8}\)
\(Q=\dfrac{2^{14}.5^8\left(2+5\right)}{2^{16}.5^7+2^{16}.5^8}\)
\(Q=\dfrac{2^{14}.5^8.7}{2^{16}.5^7\left(1+5\right)}\)
\(Q=\dfrac{5.7}{2^2.6}=\dfrac{35}{24}\)
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\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{\left(2^3\right)^5.5^8+\left(-2\right)^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{15}.5^8-2^{14}.5^9}{2^{16}.5^7\left(1+5\right)}\)
\(=\frac{2^{14}.5^8\left(2-5\right)}{2^{16}.5^7.\left(1+5\right)}\)
\(=\frac{2^{14}.5^8.\left(-3\right)}{2^{16}.5^7.6}\)
\(=\frac{-5}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)
\(=\frac{5-2^4.10}{2}\)
\(=5-8.10\)
\(=5-80\)
\(=-75\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{9}{14}-\dfrac{5}{8}+\dfrac{3}{7}=\dfrac{36}{56}-\dfrac{35}{56}+\dfrac{24}{56}=\dfrac{25}{56}\)
\(\dfrac{2}{9}-\dfrac{7}{10}+\dfrac{4}{5}=\dfrac{20}{90}-\dfrac{63}{90}+\dfrac{72}{90}=\dfrac{29}{90}\)
\(\dfrac{8}{19}+\dfrac{7}{5}-\dfrac{9}{20}+\dfrac{11}{12}=\dfrac{8}{19}+\dfrac{84}{60}-\dfrac{27}{60}+\dfrac{55}{60}=\dfrac{8}{19}+\dfrac{112}{60}=\dfrac{8}{19}+\dfrac{28}{15}=\dfrac{652}{285}\)
Đáp án là:-1.6384x10^20
Tổng trên có 10 số hạng nhóm 2 số vào một nhóm ta được:
A = (2+22)+(23+24)+...+(29+210)
A = 2(1+2) + 23(1+2)+...+29(1+2)
A = 2.3+23.3+...+29.3
A = 3(2+23+...+29) chia hết cho 3
Vậy tổng A chia hét cho 3