Giải:
a) \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow2A=3\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}\)
\(\Leftrightarrow A=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
b) Để B nguyên thì:
\(\dfrac{x+3}{x+2}\in Z\)
\(\Leftrightarrow x+3⋮x+2\)
\(\Leftrightarrow x+2+1⋮x+2\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\) (thõa mãn)
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