Câu 1:
\(A\in Z\Rightarrow6n-1⋮3n+2\)
\(\Rightarrow6n+4-5⋮3n+2\)
\(\Rightarrow2\left(3n+2\right)-5⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
đến đây tự lm nốt nhé
1. Để A có giá trị nguyên thì \(6n-1⋮3n+2\)
Ta có: \(\left\{{}\begin{matrix}6n-1⋮3n+2\\3n+2⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\2\left(3n+2\right)⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n+4⋮3n+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n-1⋮3n+2\\6n-1+5⋮3n+2\end{matrix}\right.\)
\(\Rightarrow\left(6n-1+5\right)-\left(6n-1\right)⋮3n+2\)
\(\Rightarrow5⋮3n+2\)
\(\Rightarrow3n+2\inƯ\left(5\right)\)
\(\Rightarrow3n+2\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow3n\in\left\{-7;\pm3;-1;\right\}\)
\(\Rightarrow n\in\left\{\pm1\right\}\)
Vậy để \(A\in Z\) thì n nhận các giá trị là: \(\pm1\)
2. Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
Ta có: \(2B=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(2B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2B-B=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(B=\dfrac{1}{2}-\dfrac{1}{2^{100}}\)
\(\Rightarrow B< \dfrac{1}{2}< \dfrac{2}{2}=1\)
\(\Rightarrow B< 1\left(đpcm\right)\)