a) Ta có: \(\left|1-x\right|=\dfrac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}1-x=\dfrac{1}{4}\\1-x=-\dfrac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1-\dfrac{1}{4}\\x=1-\left(-\dfrac{1}{4}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)

b) Vì \(\dfrac{x+1}{3}=\dfrac{x}{2}\Leftrightarrow2\left(x+1\right)=3x\)

\(\Leftrightarrow2x+2=3x\)

\(\Leftrightarrow x=3x-2x\)

\(\Leftrightarrow x=2\)

c) Ta có: \(\dfrac{2x}{3}-\dfrac{x}{4}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{8x}{12}-\dfrac{3x}{12}=\dfrac{10}{12}\)

\(\Rightarrow\dfrac{8x-3x}{12}=\dfrac{10}{12}\)

\(\Rightarrow\dfrac{x\left(8-3\right)}{12}=\dfrac{10}{12}\)

\(\Rightarrow\dfrac{5x}{12}=\dfrac{10}{12}\Rightarrow5x=10\Rightarrow x=2\)

Vậy \(x=2\)

b.Ta có: Đặt \(A=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(B=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)\(\Rightarrow B=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow B=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow B=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(\Rightarrow B=\dfrac{1}{25}+\dfrac{1}{26}+...+\dfrac{1}{50}=A\)

\(\Rightarrow B=A\left(đpcm\right)\)

a. Ta có: \(\dfrac{1}{21}>\dfrac{1}{40};\dfrac{1}{22}>\dfrac{1}{40};...;\dfrac{1}{40}=\dfrac{1}{40}\)

\(\Rightarrow\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}\)(20 số hạng vì A có 20 số hạng)

\(\Rightarrow A>\dfrac{1}{40}.20\)

\(\Rightarrow A>\dfrac{1}{2}\left(1\right)\)

Ta lại có: \(\dfrac{1}{21}< \dfrac{1}{20};\dfrac{1}{22}< \dfrac{1}{20};...;\dfrac{1}{40}< \dfrac{1}{20}\)

\(\Rightarrow\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{40}< \dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (20 số hạng)

\(\Rightarrow A< \dfrac{1}{20}.20\)

\(\Rightarrow A< 1\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\) ta suy ra \(\dfrac{1}{2}< A< 1\)

2. Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

Ta có: \(2B=2\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(2B-B=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(B=\dfrac{1}{2}-\dfrac{1}{2^{100}}\)

\(\Rightarrow B< \dfrac{1}{2}< \dfrac{2}{2}=1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

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: \(\pm1\)