1a.
\(A=\dfrac{2^2.3^2.2^{14}.3^{29}-2.3.2^5.7^5.2^{12}}{2.3^3.2^{14}.3^{14}.3^{14}-3.2^2.2^{15}.7^5}=\dfrac{2^{16}.3^{31}-3.2^{18}.7^5}{2^{15}.3^{31}-3.2^{17}.7^5}\)
\(A=\dfrac{2^{16}.3\left(3^{30}-2^2.7^5\right)}{2^{15}.3\left(3^{30}-2^2.7^5\right)}=\dfrac{2^{16}.3}{2^{15}.3}=2\)
1b.
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}+\dfrac{1}{2^{2019}}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}\)
\(\Rightarrow2B-B=1-\dfrac{1}{2^{2019}}\)
\(\Leftrightarrow B=1-\dfrac{1}{2^{2019}}\)
Vậy:
\(A=\left(1-\dfrac{1}{2^{2019}}\right):\left(1-\dfrac{1}{2^{2019}}\right)=1\)
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