Bài 1:
a, \(A=3^{100}+3^{99}+...+3+1\)
\(\Rightarrow3A=3^{101}+3^{100}+...+3^2+3\)
\(\Rightarrow3A-A=\left(3^{101}+3^{100}+...+3^2+3\right)-\left(3^{100}+3^{99}+...+3+1\right)\)
\(\Rightarrow2A=3^{101}+1\Rightarrow A=\dfrac{3^{101}+1}{2}\)
b, \(B=\dfrac{15^9.2^{18}.9^8}{3^{15}.4^8.25^4}=\dfrac{3^9.5^9.2^{18}.3^{16}}{3^{15}.2^{16}.5^8}\)
\(=3^{10}.5.2^2=472392\)
c, \(C=\dfrac{2^{10}.10^{17}.7^9}{5^{15}.14^9.64^9}=\dfrac{2^{10}.2^{17}.5^{17}.7^9}{5^{15}.2^9.7^9.2^{54}}\)
\(=\dfrac{5^2}{2^{36}}\)
Chúc bạn học tốt!!!
1.
\(A=3^{100}+3^{99}+3^{98}+...+3^2+3+1\\ A=\dfrac{3-1}{2}\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)}{2}\\ =\dfrac{3^{101}-3^{100}+3^{100}-3^{99}+...+3^2-3+3-1}{2}\\ =\dfrac{3^{101}-1}{2}\)
\(B=\dfrac{15^9\cdot2^{18}\cdot9^8}{3^{15}\cdot4^8\cdot25^4}\\ =\dfrac{\left(3\cdot5\right)^9\cdot2^{18}\cdot\left(3^2\right)^8}{3^{15}\cdot\left(2^2\right)^8\cdot\left(5^2\right)^4}\\ =\dfrac{3^9\cdot5^9\cdot2^{18}\cdot3^{16}}{3^{15}\cdot2^{16}\cdot5^8}\\ =\dfrac{3^9\cdot5\cdot2^2\cdot3}{1\cdot1\cdot1}\\ =3^{10}\cdot5\cdot2^2\\ =59049\cdot5\cdot4\\ =59049\cdot\left(5\cdot4\right)\\ =59049\cdot20\\ =1180980\)
\(C=\dfrac{2^{10}\cdot10^{17}\cdot7^9}{5^{15}\cdot14^9\cdot64^9}\\ =\dfrac{2^{10}\cdot\left(2\cdot5\right)^{17}\cdot7^9}{5^{15}\cdot\left(2\cdot7\right)^9\cdot\left(2^6\right)^9}\\ =\dfrac{2^{10}\cdot2^{17}\cdot5^{17}\cdot7^9}{5^{15}\cdot2^9\cdot7^9\cdot2^{54}}\\ =\dfrac{2\cdot1\cdot5^2\cdot1}{1\cdot1\cdot1\cdot2^{37}}\\ =\dfrac{5^2}{2^{36}}\\ =\dfrac{25}{2^{36}}\)
