=2700

=> 2700

nhớ cho cả hướng dẫn giải

a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)

b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)

1) \(=\frac{6^5.5^3\left(1+5\right)}{6^5.5^3.3}=\frac{6}{3}=2\)

2)

\(2B=2+2^2+2^3+...+2^{101}\)

\(2B-B=B=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)=2^{101}-1\)

(0.6)^5/(0.2)^6 
= [(0.2)^5. 3^5]/(0.2)^6 
= 3^5/0.2 
=243/0.2 
= 1215 

6^3+3.6^2+3^3/-13 
= 2^3.3^3+3.2^2.3^2+3^3/-13 
=3^3(2^3+2^2+1)/-13 
=3^3.13/-13 
=-27

\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{\left(3\cdot2\right)^3+3\cdot\left(3\cdot2\right)^2+3^3}{-13}=\frac{3^3\cdot2^3+3\cdot3^2\cdot2^2+3^3}{-13}=\frac{3^3\cdot\left(2^3+2^2+1\right)}{-13}=\frac{27\cdot13}{-13}=-27\)

a)

\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).