Question

Tìm x

\(\frac{-3}{29}+\frac{-7}{29}\le\frac{x}{29}\le\frac{-3}{29}-\frac{5}{29}\)

Tìm \(n\in Z\)để \(\frac{n^2+n-5}{n+2}\)là số nguyên

Tính \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

 

 

Answer 1

\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)

\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)

\(\Rightarrow x=\left\{-8;-9;-10\right\}\)

\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

             \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)

\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)