\(\dfrac{x-1}{9}+\dfrac{x-2}{8}+\dfrac{x-3}{7}=3\)

\(\Leftrightarrow\left(\dfrac{x-1}{9}-1\right)+\left(\dfrac{x-2}{8}-1\right)+\left(\dfrac{x-3}{7}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-10}{9}+\dfrac{x-10}{8}+\dfrac{x-10}{7}=0\)

\(\Leftrightarrow\left(x-10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

Ta có: \(\dfrac{n-8}{n-3}=\dfrac{n-3-5}{n-3}=1-\dfrac{5}{n-3}\)

\(\Rightarrow n-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

TH1:\(n-3=1\Rightarrow n=4\Rightarrow A=-4\)

TH2: \(n-3=-1\Rightarrow n=2\Rightarrow A=6\)

TH3:\(n-3=5\Rightarrow n=8\Rightarrow A=0\)

TH4:\(n-3=-5\Rightarrow n=-2\Rightarrow A=2\)

Vậy A đạt giá trị lớn nhất khi n =2

Ta có: \(2A=26+46+...+206\)

\(=23+3+43+3+...+203+3\)

=\(B+30\)

\(\Rightarrow B=2A-30\)

\(=2.580-30\)

\(=1160-30=1130\)

Vậy B = 1130

Chứng minh rằng với mỗi số nguyên dương n thì:

\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)

\(\left|x-1\right|+\left|x-2\right|\)

=\(\left|1-x\right|+\left|x-2\right|\ge\left|1-x+x-2\right|=\left|-1\right|=1\)

\(\Rightarrow\left|x-1\right|+\left|x-2\right|\ge1\)

Dấu "=" xảy ra khi \(\left(1-x\right)\left(x-2\right)\ge0\)

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}1-x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)(không có số nào thảo mãn)

\(\Leftrightarrow1\le x\le2\)

Vậy GTNN của\(\left|x-1\right|+\left|x-2\right|\) là 1 khi \(1\le x\le2\)