b:\(=\dfrac{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

a: \(=\dfrac{5^5\cdot4^5\cdot5^{10}}{10^{20}}=\dfrac{5^{15}\cdot2^{10}}{5^{20}\cdot2^{20}}=\dfrac{1}{5^5\cdot2^{10}}\)

b) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\dfrac{3^3.\left(8+4+1\right)}{-13}=\dfrac{3^3.13}{-13}=\dfrac{27}{-1}=-27\)

a)\(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=\dfrac{5^5}{1}=3125\)

a, 20⁵.5¹⁰/100⁵

=20⁵.5⁵.5⁵/100⁵

=100⁵.5⁵/100⁵

=5⁵

=3125

b, (0,9)⁵/(0,3)⁶

=(0,3.3)⁵/0,3⁶

=0,5⁵.3⁵/0,3⁶

=3⁵/0,3

=810

c, 6³+3.6²+3³/-13

=(2.3)³+3.(3.2)²+3³/-13

=2³.3³+3.3².2²+3³/-13

=3³.2³+3³.2²+3³/-13

=3³(2³+2²+1)/-13

=27.13/-13

=-27

d, 4⁶.9⁵+6⁹.120/8⁴.3¹²-6¹¹

=(2²)⁶.(3²)⁵+(2.3)⁹.3.2³.5/(2³)⁴.3¹²-(2.3)¹¹

=2¹².3¹⁰+2⁹.3⁹.3.2³.5/2¹².3¹²-2¹¹.3¹¹

=2¹².3¹⁰+2¹².3¹⁰.5/2¹¹.3¹¹.2.3-2¹¹.3¹¹

=2¹².3¹⁰.(1+5)/2¹¹.3¹¹(6-1)

=2¹².3¹⁰.6/2¹¹.3¹¹.5

=2.6/3.5

=12/15

=4/5

a. \(\frac{20^5.5^{10}}{100^5}\)= \(\frac{20^5.5^{10}}{20^5.5^5}\)= \(5^5\)=\(3125\)

b. \(\frac{0,9^5}{0,3^6}\)= \(\frac{0,9^5}{0,3^5.0,3}\)= \(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)= \(243.\frac{1}{0,3}\)= \(810\)

c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)

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

Tính bình thường dc ko bn

6^2.(6+3)+3^3/-13=36.3^2+3^3/-13=3^2(36+3)/-13=9.39/-13=-27

2^10.3^8-2.3^9.2^9 / 2^10.3^8+3^8.2^8.2^2.5

=2^10.3^8(1-3) / 2^10.3^8(1+5)

=-2/6=-1/3

(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/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)