\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)

=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)=     0

\(B=-\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)

\(=-\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)

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

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

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)

                                Bài giải

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

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)

\(D=\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+3^9.2^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\\ =\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-6^{12}-6^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{6^{11}\left(-6-1\right)}=\dfrac{2^{12}.3^{10}.6}{6^{11}.-7}=\dfrac{2^{12}.3^{10}.2.3}{6^{11}.-7}\\ =\dfrac{2^{13}.3^{11}}{6^{11}.-7}=\dfrac{2^{11}.3^{11}.2^2}{6^{11}.-7}=\dfrac{6^{11}.4}{6^{11}.-7}=-\dfrac{4}{7}\)

D=4^6.9^5+6^9.120/-8^4.3^12-6^11
= 2^12.3^10+6^10.20/(-2)^12.3^12-6^11
= 20/3^2.6=10/27

a)  \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{5^{10}.9^{10}.5^{20}}{5^{15}.5^{15}.3^{15}}=\frac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)

b) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)

\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=-3^3=-27\)

c) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)

\(=\frac{2.6}{3.5}=\frac{4}{5}\)

d) \(\frac{3.11+42}{5^3}=\frac{33+42}{5^3}=\frac{75}{5^3}=\frac{5^2.3}{5^3}=\frac{3}{5}\)

Câu a là 243

Câu b là -27

Câu c là 4 phần 5

câu d là 3 phần 5

GIÚP MÌNH VỚI MN ƠI

\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)