\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)

\(=\dfrac{140}{27}+\dfrac{7}{23}+\dfrac{1}{2}-\dfrac{5}{27}+\dfrac{16}{23}\)

\(=\dfrac{140}{27}-\dfrac{5}{27}+\dfrac{7}{23}+\dfrac{16}{23}+\dfrac{1}{2}\)

\(=5+1+\dfrac{1}{2}=6+\dfrac{1}{2}=\dfrac{13}{2}\)

a) https://hoc247.net/hoi-dap/toan-7/thuc-hien-phep-tinh-5-5-27-7-23-0-5-5-27-16-23-faq218258.html

b) https://hoidap247.com/cau-hoi/1507941

b) \(\dfrac{10^9.49^4}{14^8.25^5}=\dfrac{\left(2.5\right)^9.\left(7^2\right)^4}{\left(7.2\right)^8.\left(5^2\right)^5}=\dfrac{2^9.5^9.7^6}{7^8.2^8.5^{10}}=\dfrac{2}{7.7.5}=\dfrac{2}{245}\)

\(a,=\left(\dfrac{140}{27}+\dfrac{1}{2}\cdot\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)\\ =\left(\dfrac{140}{27}+\dfrac{5}{54}\right)+1=\dfrac{285}{54}+1=\dfrac{339}{54}=\dfrac{113}{18}\\ b,=\dfrac{2^9\cdot5^9\cdot7^8}{7^8\cdot2^8\cdot5^{10}}=\dfrac{2}{5}\)

Dấu " / " là phân số nhé :

a) 5 5/27 + 7/23 + 0,5 - 5/27 + 16/23

= 140/27 + 7/23 + 1/2 - 5/27 + 16/23

= ( 140/27 - 5/27 ) + ( 7/23 + 16/23 ) + 1/2

= 5 + 1 + 1/2

= 6 + 1/2

= 13/2

b) 3/8 . 27 1/5 - 51 .1/5 . 3/8 + 19

= 3/8 . 136/5 - 51 . 1/5 . 3/8 + 19

= 3/8 . ( 136/5 - 1/5 ) - 51 + 19

= 3/8 . 27 - 51 + 19

= 81/8 - 52 + 19

= -183/8

Sorry nha mik sửa lại con B

b) 3/8 . 27 1/5 - 51 . 1/5 . 3/8 + 19

= 3/8 . 136/5 - 51/5 . 3/8 + 19

= 3/8 . ( 136/5 - 51/5 ) + 19

= 3/8 . 17 + 19

= 51/8 + 19

= 203/8

Ai giúp mình làm bước tiếp theo với ạ   

a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)

b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)

c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)

e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)

\(1:5\dfrac{25}{7}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)

\(=\dfrac{60}{7}+\left(\dfrac{27}{23}+\dfrac{16}{23}\right)+0,5-\dfrac{5}{27}\)

\(=\dfrac{60}{7}+\dfrac{43}{23}+0,5-\dfrac{5}{27}\)

\(=\dfrac{1681}{161}+\dfrac{1}{2}-\dfrac{5}{27}\)

\(=\dfrac{3523}{322}-\dfrac{5}{27}\)

\(=10,7558086\)

\(2:\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{15}-\left(-\dfrac{1}{5}\right):\left(-3\right)\)

\(=-1:\left(-0,5\right)+\dfrac{1}{15}-\dfrac{1}{15}\)

\(=2+0=2\)

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

a) Ta có: \(\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-\dfrac{19}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\dfrac{-9}{16}\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}\cdot\dfrac{-16}{9}\)

\(=\dfrac{-112}{144}=\dfrac{-7}{9}\)

b) Ta có: \(17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)

\(=17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}\cdot\dfrac{4}{27}+350\%\)

\(=\dfrac{4}{27}\left(17+\dfrac{6}{11}-8-\dfrac{6}{11}\right)+\dfrac{7}{2}\)

\(=\dfrac{4}{27}\cdot9+\dfrac{7}{2}\)

\(=\dfrac{4}{3}+\dfrac{7}{2}=\dfrac{8}{6}+\dfrac{21}{6}=\dfrac{29}{6}\)