B = \(\frac{\left(2^3\right)^7.3^9}{2^9.3^9.2^{11}}\)

B = \(\frac{2^{21}}{2^{20}}\)

B = \(2\)

\(B=\frac{8^7.3^9}{6^9.2^{11}}=\frac{\left(2^3\right)^7.3^9}{2^9.3^9.2^{11}}=\frac{2^{21}.3^9}{2^{20}.3^9.}=2\)

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

Thay x=-8 và y=6 cào C ta được:

\(C=\dfrac{\left(-8\right)^3}{2}+\dfrac{\left(-8\right)^2.6}{4}+\dfrac{\left(-8\right).6^2}{6}+\dfrac{6^3}{27}\)\(=\dfrac{-512}{2}+\dfrac{384}{4}-\dfrac{288}{6}+\dfrac{216}{27}\)\(=-256+96-48+8=-200\)

\(C=x^2\left(\dfrac{x}{2}+\dfrac{y}{4}\right)+y^2\left(\dfrac{x}{6}+\dfrac{y}{27}\right)=\left(-8\right)^2\left(-\dfrac{8}{2}+\dfrac{6}{4}\right)+6^2\left(-\dfrac{8}{6}+\dfrac{6}{27}\right)=-200\)

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)

\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)

\(=\dfrac{3}{4}-\dfrac{5}{12}\)

\(=\dfrac{4}{12}=\dfrac{1}{3}\)

B = \(\left(-1\dfrac{1}{6}\right)\) : \(\left(\dfrac{-10}{3}+\dfrac{9}{4}\right)\) - \(\left(-\dfrac{3}{8}\right)\) : \(\left(8-\dfrac{51}{8}\right)\)

B = \(\dfrac{-7}{6}\) : \(\dfrac{-13}{12}\) - \(\left(-\dfrac{3}{8}\right)\) : \(\dfrac{13}{8}\)

B = \(\dfrac{14}{13}\) - \(\dfrac{-3}{13}\)

B = \(\dfrac{17}{13}\)