Bài 1:

a) \(a=2\cdot3\cdot5\cdot43\)

\(b=7200=2^5\cdot3^2\cdot5^2\)

\(c-4680=2^3\cdot3^2\cdot5\cdot13\)

b) \(\dfrac{8440}{5910}=\dfrac{8440:10}{5910:10}=\dfrac{844}{591}\)

\(\dfrac{1245}{3450}=\dfrac{1245:15}{3450:15}=\dfrac{83}{230}\)

Bài 2:

a) Ước nguyên tố của 140 là:

\(ƯNT\left(140\right)=\left\{2;5;7\right\}\)

Ước nguyên tố của 138 là:
\(ƯNT\left(138\right)=\left\{3;23;2\right\}\)

b) \(A=\dfrac{2^{10}+4^6}{8^4}\)

\(A=\dfrac{2^{10}+2^{12}}{2^{12}}\)

\(A=\dfrac{2^{10}\cdot\left(1+2^2\right)}{2^{12}}\)

\(A=\dfrac{1+4}{2^2}\)

\(A=\dfrac{5}{4}\)

\(B=\dfrac{6^{10}+15\cdot2^{10}\cdot3^9}{12\cdot8^3\cdot27^3}\)

\(B=\dfrac{2^{10}\cdot3^{10}+5\cdot2^{10}\cdot3^{10}}{2^{11}\cdot3^{10}}\)

\(B=\dfrac{2^{10}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{10}}\)

\(B=\dfrac{1+5}{2}\)

\(B=3\)

a: \(=\dfrac{3^6\cdot2^{21}}{5^{18}\cdot7^9\cdot7^3}=\dfrac{3^6\cdot2^{21}}{5^{18}\cdot7^{12}}\)

b: \(=\dfrac{3^{10}\cdot3^7\cdot2^7\cdot2^2}{2^9\cdot5^9\cdot5^8}=\dfrac{3^{17}}{5^{17}}\)

\(A=\frac{10^6+7}{10^6}\)\(=\frac{10^6}{10^6}+\frac{7}{10^6}=1+\frac{7}{10^6}\)

\(A=4\sqrt{32}+2\sqrt{50}-8\sqrt{2}-2\sqrt{98}\)

\(=4\sqrt{16.2}+2\sqrt{25.2}-8\sqrt{2}-2\sqrt{49.2}\)

\(=16\sqrt{2}+10\sqrt{2}-8\sqrt{2}-14\sqrt{2}=4\sqrt{2}\)

\(B=\frac{1}{\sqrt{6}+\sqrt{10}}-\frac{1}{\sqrt{6}-\sqrt{10}}\)

\(=\frac{\sqrt{10}-\sqrt{6}}{\left(\sqrt{6}+\sqrt{10}\right)\left(\sqrt{10}-\sqrt{6}\right)}+\frac{\sqrt{6}+\sqrt{10}}{\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{6}+\sqrt{10}\right)}\)

\(=\frac{\sqrt{10}-\sqrt{6}}{4}+\frac{\sqrt{10}+\sqrt{6}}{4}\)

\(=\frac{2\sqrt{10}}{4}=\frac{\sqrt{10}}{2}=\sqrt{2,5}\)

A=\(4\sqrt{2}\)

B=\(\frac{\sqrt{10}}{2}\)

\(A=\frac{10^6+7}{10^6}=1+\frac{7}{10^6}\)

\(B=\frac{10^8-2}{10^8+5}=\frac{10^8+5-7}{10^8+5}=1-\frac{7}{10^8}\)

\(A=\frac{4^6.9^5+69.120}{8^4.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3^2.115}{2^{12}.3^{12}+\left(2.3\right)^{11}}=\frac{3^2.2^3\left(115+2^9.3^8\right)}{6^{11}\left(6+1\right)}=\frac{115+2^9.3^8}{6^8.3.7}\)

\(B=\frac{10^4+5.10^3+5^4}{25}=\frac{\left(10^2\right)^2+2.5^2.10^2+\left(5^2\right)^2}{25}=\frac{\left(10^2+5^2\right)^2}{25}=\frac{125^2}{25}=\frac{25.625}{25}=625\)

\(C=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{10}.2^{10}+4^{10}}{4^4.4^7+4^4.2^4}=\frac{4^{10}\left(2^{10}+1\right)}{4^4.2^4\left(2^{10}+1\right)}=\frac{4^6}{2^4}=256\)

a)\(=\frac{3^7}{2^2.3^4}=\frac{3^3}{2^2}=\frac{27}{4}\)

b)\(=\frac{2^{20}+2^{12}}{2^{10}+2^{18}}=\frac{2^{12}\left(2^8+1\right)}{2^{10}\left(1+2^8\right)}=\frac{2^{12}}{2^{10}}=2^2=4\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)

Ta có:\

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(A=-\frac{2}{6}=-\frac{1}{3}\)

Ta có : 

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.4.5}\)

\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(A=\frac{1-3}{1+5}\)

\(A=\frac{-2}{6}\)

\(A=\frac{-1}{3}\)

Chúc bạn học tốt ~