Giải:

a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\) 

\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\) 

\(=\dfrac{136}{9}-\dfrac{11}{2}\) 

\(=\dfrac{173}{18}\) 

b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\) 

\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\) 

\(=\dfrac{13}{9}.\dfrac{3}{4}\) 

\(=\dfrac{13}{12}\) 

c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\) 

\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\) 

\(=1+1\) 

\(=2\)

d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\) 

\(=\dfrac{53}{10}\) 

e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-2}{5}\) 

\(=1\)

a) \(\sqrt{4}+\sqrt{4}\)

\(=2+2\)

\(=4\)

b) \(\sqrt{9+9+9+9+9.5}+\sqrt{81}\)

\(=9+9\)

\(=18\)

c) \(\sqrt{15}+\sqrt{15}=2\sqrt{15}\)

a) \(\sqrt{4}\)+ \(\sqrt{4}\)= 2 + 2 = 4

b) \(\sqrt{9+9+9+9+9.5}\)+ \(\sqrt{81}\)= \(\sqrt{81}\)+\(\sqrt{81}\)= 9 + 9 = 18

c) \(\sqrt{15}\)+\(\sqrt{15}\)= 7,745966692

3².5²⁰.(3³)⁹-3.(3²)¹⁵.(5²)⁹

7.3²⁹.(5³)

1. \(\dfrac{9.5^{20}.27^9-3.9^{15}:25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)

\(=\dfrac{3^2.5^{20}.3^{27}-3.3^{30}.5^{18}}{7.3^{29}.5^{18}-3^{10}.3^{19}.5^{19}}\)

\(=\dfrac{3^{29}.5^{20}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}\)

\(=\dfrac{3^{28}.5^{18}.\left(5^2-3^2\right)}{3^{29}.5^{18}.\left(7-5\right)}\)

\(=\dfrac{5^2-3^2}{7-5}=\dfrac{16}{2}=8\)

2.\(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{-4}\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}^2\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

3.\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(=\dfrac{1-3}{1+5}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

Chúc học tốt!!

\(=\dfrac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{13}-3^{29}\cdot5^{19}}=\dfrac{3^{29}\cdot5^{20}-5^{18}\cdot3^{31}}{3^{29}\cdot5^{13}\cdot7-3^{29}\cdot5^{19}}\)

\(=\dfrac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{13}\left(7-5^6\right)}=5^5\cdot\dfrac{4^2}{7-5^6}\)

\(-\dfrac{5}{9}\times\dfrac{5}{11}+\dfrac{6}{9}\times\dfrac{5}{11}+\dfrac{5}{9}\)

\(=\dfrac{5}{11}\times\left(-\dfrac{5}{9}+\dfrac{6}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{5}{11}\times\dfrac{6}{9}\)

\(=\dfrac{30}{99}\)

\(=\dfrac{10}{33}\)

Cho mình sửa lại ạ:

\(-\dfrac{5}{9}\times\dfrac{5}{11}+\dfrac{6}{9}\times\dfrac{5}{11}+\dfrac{5}{9}\)

\(=\dfrac{5}{11}\times\left(-\dfrac{5}{9}+\dfrac{6}{9}\right)+\dfrac{5}{9}\)

\(=\dfrac{5}{11}\times\dfrac{1}{9}+\dfrac{5}{9}\)

\(=\dfrac{5}{99}+\dfrac{5}{9}\)

\(=\dfrac{60}{99}\)

\(=\dfrac{20}{33}\)

Bài làm

     \(\frac{2}{3}.\frac{7}{15}-\frac{2}{3}.\frac{5}{27}\)

\(=\frac{2}{3}.\left(\frac{7}{15}-\frac{5}{27}\right)\)

\(=\frac{2}{3}.\left(\frac{63}{135}-\frac{25}{135}\right)\)

\(=\frac{2}{3}.\frac{38}{135}\)

\(=\frac{76}{405}\)

~ Nếu rút gọn đc rút hộ ~

        \(-\frac{4}{9}.\frac{7}{15}+\frac{4}{9}.\frac{5}{27}\)

\(=-\frac{4}{9}.\frac{7}{15}-\left(-\frac{4}{9}\right).\frac{5}{27}\)

\(=-\frac{4}{9}.\left(\frac{7}{15}-\frac{5}{27}\right)\)

\(=-\frac{4}{9}.\left(\frac{63}{135}-\frac{25}{135}\right)\)

\(=-\frac{4}{9}.\frac{38}{135}\)

\(=\frac{152}{1215}\)

~ Rút gọn đc thì rút ~
# Chúc bạn học tốt #

\(\frac{\frac{2}{3}.\frac{7}{15}-\frac{2}{3}.\frac{5}{27}}{\frac{-4}{9}.\frac{7}{15}+\frac{4}{9}.\frac{5}{27}}=\frac{\frac{2}{3}.\left(\frac{7}{15}-\frac{5}{27}\right)}{\frac{4}{9}\left(\frac{-7}{15}+\frac{5}{27}\right)}=\frac{\frac{2}{3}}{\frac{4}{9}}=\frac{2:3}{4:9}=\frac{2.\frac{1}{3}}{4.\frac{1}{9}}=\frac{1.\frac{1}{3}}{2.\frac{1}{9}}=\frac{1}{6}\)

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