Question

giúp mk với ạ

Answer 1

a, \(S=\dfrac{3\cdot\left(\dfrac{1}{13}-0,2+\dfrac{1}{7}+0.25\right)}{11\cdot\left(\dfrac{1}{7}-0.2+\dfrac{1}{13}+0.25\right)}=\dfrac{3}{11}\)

b, Ta có: \(\dfrac{1}{3}A=-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...+\dfrac{1}{3^{101}}\)

\(\Rightarrow A+\dfrac{1}{3}A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)

\(\Rightarrow\dfrac{4}{3}A=-\dfrac{1}{3}+\dfrac{1}{3^{101}}\)

hay \(A=-\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{3^{100}}=\dfrac{1}{4}\cdot\left(\dfrac{1}{3^{100}}-1\right)\)

\(\Rightarrow B=4\left|A\right|+\dfrac{1}{3^{100}}=\left|4A\right|+\dfrac{1}{3^{100}}=\left|\dfrac{1}{3^{100}}-1\right|+\dfrac{1}{3^{100}}=1-\dfrac{1}{3^{100}}+\dfrac{1}{3^{100}}=1\)

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