\(A=\dfrac{1}{7}\left(\dfrac{555}{777}+\dfrac{4444}{12221}+\dfrac{33333}{244442}+\dfrac{11}{30}+\dfrac{13}{30}\right)\)

\(=\dfrac{1}{7}\left(\dfrac{5}{7}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{23}{30}\right)\\ =\dfrac{1}{7}\left(\dfrac{5}{7}+\dfrac{1}{2}+\dfrac{23}{30}\right)\\ =\dfrac{1}{7}\cdot\dfrac{208}{105}\\ =\dfrac{208}{735}\)

\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\\ =\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-\dfrac{5}{7}+\left(-\dfrac{20}{41}-\dfrac{21}{41}\right)\\ =1-\dfrac{5}{7}-1\\ =-\dfrac{5}{7}\)

\(\dfrac{5}{9}\cdot\dfrac{14}{17}+\dfrac{1}{17}\cdot\dfrac{5}{9}+\dfrac{2}{9}\cdot\dfrac{5}{17}\\ =\dfrac{5}{9}\cdot\dfrac{14}{17}+\dfrac{1}{17}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{2}{17}\\ =\dfrac{5}{9}\left(\dfrac{14}{17}+\dfrac{1}{17}+\dfrac{2}{17}\right)\\ =\dfrac{5}{9}\cdot1\\ =\dfrac{5}{9}\)

\(\dfrac{-2}{5}\cdot\dfrac{4}{7}+\dfrac{-3}{5}\cdot\dfrac{2}{7}+\dfrac{-3}{5}\)

\(=-\dfrac{8}{35}-\dfrac{6}{35}-\dfrac{21}{35}\\ =-\dfrac{35}{35}\\ =-1\)

\(3\cdot2\cdot x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{3}=\dfrac{7}{20}\)

\(6x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)\cdot\dfrac{3}{11}=\dfrac{7}{20}\)

\(6x-\dfrac{22}{15}\cdot\dfrac{3}{11}=\dfrac{7}{20}\\ 6x-\dfrac{2}{5}=\dfrac{7}{20}\\ 6x=\dfrac{7}{20}+\dfrac{2}{5}\\ 6x=\dfrac{3}{4}\\ x=\dfrac{1}{8}\)

Để \(A=\dfrac{1}{x^2-x+1}\) có giá trị lớn nhất thì \(x^2-x+1\) phải nhỏ nhất

Mà :

\(x^2-x+1\\ =\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Vậy \(Min_A=\dfrac{1}{\left(\dfrac{3}{4}\right)}=\dfrac{4}{3}\) khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

\(\left(x-3\right)\left(x-5\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-3< 0\\x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3>0\\x-5< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x>5\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x< 5\end{matrix}\right.\end{matrix}\right.\)

Vậy 3<x<5 là giá trị cần tìm

\(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\\ =\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{-3\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\\ =\dfrac{2}{-3}\\ =-\dfrac{2}{3}\)

\(A=\left(-1,5\right)^2\cdot2\dfrac{2}{3}-\dfrac{1}{6}+\left(\dfrac{4}{7}-\dfrac{2}{5}\right):1\dfrac{1}{35}\)

\(=\left(-\dfrac{3}{2}\right)^2\cdot\dfrac{8}{3}-\dfrac{1}{6}+\left(\dfrac{20}{35}-\dfrac{14}{35}\right):\dfrac{36}{35}\\ =\dfrac{9}{4}\cdot\dfrac{8}{3}-\dfrac{1}{6}+\dfrac{6}{35}\cdot\dfrac{35}{36}\\ =6-\dfrac{1}{6}+\dfrac{1}{6}\\ =6\)

a) \(Q\left(x\right)=x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\\\Rightarrow \left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

b) \(N\left(x\right)=x^2-4=0\)

\(\Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)

c) \(H\left(x\right)=8-3x=0\)

\(\Leftrightarrow-3x=-8\\ \Leftrightarrow x=\dfrac{8}{3}\)

d) \(g\left(x\right)=x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)