\(a,A=\frac{3}{5}.\frac{5}{4}-\frac{3}{5}.\frac{1}{4}\)

\(=\frac{3}{5}.\left(\frac{5}{4}-\frac{1}{4}\right)\)

\(=\frac{3}{5}.1=\frac{3}{5}\)

\(b,B=6,3+\left(-6,3\right)+4,9\)

\(=0+4,9=4,9\)

b, \(A=\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)

\(A=13+...+3^9.13=13\left(1+...+3^9\right)⋮13\)

`Answer:`

A)  \(A=1+3+3^2+3^3+3^4+...+3^{11}\)

\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{12}\)

\(\Rightarrow3A-A=3^{12}-1\)

\(\Rightarrow A=\frac{3^{12}-1}{2}\)

B) Ta có:

 \(1+3+3^2=13⋮13\)

\(3^3+3^4+3^5=3^3.\left(1+3+3^2\right)⋮13\)

...

\(3^9+3^{10}+3^{11}=3^9.\left(1+3+3^2\right)⋮13\)

\(\Rightarrow A⋮13\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=\frac{99}{100}\)

\(\dfrac{21}{105}=\dfrac{35-x}{105}\Rightarrow35-x=21\Leftrightarrow x=14\)

\(\dfrac{2x^2}{3}=\dfrac{8}{3}\Rightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(\frac{2-x}{-27}=\frac{-3}{2-x}\)

\(\Leftrightarrow\left(2-x\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2-x\right)^2=9^2\\\left(2-x\right)^2=\left(-9\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}2-x=9\\2-x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=11\end{cases}}\)

Vậy x = -7 hoặc x = 11

`Answer:`

\(E=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)

\(=\frac{5.4}{5.\left(7.31\right)}+\frac{5.6}{5.\left(7.41\right)}+\frac{5.9}{5.\left(10.41\right)}+\frac{5.17}{5.\left(10.57\right)}\)

\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)

\(=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)

\(=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)

\(=5.\frac{26}{1767}\)

\(=\frac{130}{1767}\)