Ta có: \(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{3^{18}\cdot2^{29}\left(5\cdot2-3^2\right)}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{3^{18}\cdot2^{29}}{2^{29}\cdot3^{18}\left(5\cdot2-7\right)}\)

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

Bạn viết đề như này sao hiểu đc

a) 19 + (29 - 9*37) - (63*9 - 29*99)

= 19 + 29 - 9*37 - 63*9 + 29*99

= 19 + 29(1 + 99) - 9(37 + 63)

= 19 + 29*100 - 9*100

= 19 + 100(29 - 9)

= 19 + 100*20

= 19 + 2000 = 2019

b) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)

= \(\frac{2^6+2^5+2^4+2^3+2^2+2+1}{2^7}\)

= \(\frac{64+32+16+8+4+2+1}{128}\) = \(\frac{127}{128}\)

c) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)= \(\frac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(10-9\right)}{2^{28}.3^{18}\left(15-14\right)}=2\)

a: =-2/49-5/3+2/49=-5/3

b: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)

mai mình thi rồi, các bạn giúp mình với

Chờ tí

Câu 4: 

d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)

\(=\left(-\dfrac{2}{17}-\dfrac{15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)

\(=-1+1+\dfrac{5}{6}=\dfrac{5}{6}\)

Câu 3:

a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)

\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)

b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)

\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)

Câu 4: 

a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)

b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)

\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)

\(=-27\cdot100-23\cdot100\)

\(=100\left(-27-23\right)\)

\(=-50\cdot100=-5000\)

c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)

d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)

\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)

\(=-1+1+\dfrac{5}{6}\)

\(=\dfrac{5}{6}\)

1)C= 1/5+1/10+1/20+1/40+...+1/1280

\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

Đặt cái trong ngoặc là A ta có:\(2A=2+1+...+\frac{1}{2^7}\)

\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

\(A=2-\frac{1}{2^8}\).Thay A vào ta được:\(C=\frac{1}{5}\left(2-\frac{1}{2^8}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)

2)D= 2/1*3+2/3*5+2/5*10+2/7*9+2/9*11+2/11*18+2/13*15

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

3)E= 4/3*7+4/7*11+4/11*15+4/15*19+4/19*23+4/23*27

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

4)G= 1/2+1/6+1/12+1/20+1/30+1/42+...+1/110

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

5)H= 3/1*2+3/2*3+3/3*4+3/4*5+...+3/9*10

\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=3\left(1-\frac{1}{10}\right)\)

\(=3\times\frac{9}{10}\)

\(=\frac{27}{10}\).Lần sau bạn đăng ít một thôi nhé 

A = 2¹⁹ x 3⁹ x 12151

cho mình biết cách làm đươc ko

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^9}{2^9.3^9.2^{10}+\left(3.2^2\right)^{10}}\)

\(=\frac{2^{19}.3^9+5.2^{18}.3^{19}}{2^{19}.3^9+3^{10}.2^{20}}=\frac{2^{18}.3^9\left(2+5.3\right)}{2^{18}.3^9\left(2+3.2^2\right)}=\frac{17}{14}\)

thank kiu, very much