\(\dfrac{9}{17}x+15\dfrac{13}{17}x-20\dfrac{5}{17}x=16\)

\(\left(\dfrac{9}{17}+15\dfrac{13}{17}-20\dfrac{5}{17}\right)x=16\)

\(\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)x=16\)

\(x=16:\left(\dfrac{9}{17}+\dfrac{268}{17}-\dfrac{345}{17}\right)\)

\(x=-4\)

\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

\(\dfrac{2}{3}x+\dfrac{3}{4}=\dfrac{5}{6}\)

\(\dfrac{2}{3}.x=\dfrac{5}{6}-\dfrac{3}{4}\)

\(\dfrac{2}{3}.x=\dfrac{10}{12}-\dfrac{9}{12}\)

\(\dfrac{2}{3}.x=\dfrac{1}{12}\)

\(x=\dfrac{1}{12}:\dfrac{2}{3}\)

\(x=\dfrac{1}{12}.\dfrac{3}{2}\)

\(x=\dfrac{1}{8}\)

\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\Rightarrow\dfrac{A}{5}=\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\)

\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\)

\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}\)

\(\dfrac{B}{2}=\dfrac{7}{38.31}+\dfrac{5}{38.41}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\Rightarrow\dfrac{B}{2}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\)

\(\Rightarrow B=\dfrac{52}{31.57}\)

\(\dfrac{A}{B}=\dfrac{130}{31.51}:\dfrac{52}{31.57}=\dfrac{5}{2}\)

\(A=\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}-\dfrac{17}{72}+\dfrac{19}{90}\)

\(=\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{90}+\dfrac{2}{3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.10}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

Có \(|x-9|\ge0\forall x\Rightarrow|x-9|+10\ge10\forall x\)

Dấu "=" xảy ra khi \(x=9\)

Vậy \(minA=10\) khi \(x=9\)

\(A=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+8}=1+\dfrac{1}{2.3:2}+\dfrac{1}{3.4:2}+...+\dfrac{1}{8.9:2}\)

\(=1+2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}\right)=1+2.\left(\dfrac{1}{2}-\dfrac{1}{9}\right)=1+2.\dfrac{7}{18}=\dfrac{16}{9}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2020^2}< \dfrac{1}{2019.2020}\)

Vậy \(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2019.2020}=\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}< \dfrac{2020}{2020}=1\)