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\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{140}{27}+\dfrac{7}{23}+\dfrac{1}{2}-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{140}{27}-\dfrac{5}{27}+\dfrac{7}{23}+\dfrac{16}{23}+\dfrac{1}{2}\)
\(=5+1+\dfrac{1}{2}=6+\dfrac{1}{2}=\dfrac{13}{2}\)
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
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
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: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
\(1:5\dfrac{25}{7}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{60}{7}+\left(\dfrac{27}{23}+\dfrac{16}{23}\right)+0,5-\dfrac{5}{27}\)
\(=\dfrac{60}{7}+\dfrac{43}{23}+0,5-\dfrac{5}{27}\)
\(=\dfrac{1681}{161}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{3523}{322}-\dfrac{5}{27}\)
\(=10,7558086\)
\(2:\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{15}-\left(-\dfrac{1}{5}\right):\left(-3\right)\)
\(=-1:\left(-0,5\right)+\dfrac{1}{15}-\dfrac{1}{15}\)
\(=2+0=2\)
\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
Giải:
\(5\dfrac{27}{5}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=5+\dfrac{27}{5}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=5+\dfrac{27}{5}+\dfrac{43}{23}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{52}{5}+\dfrac{43}{23}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{1411}{115}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{2937}{230}-\dfrac{5}{27}\)
\(\approx12,58\)
Vậy ...
\(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
=\(\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
=(\(\frac{52}{5}+\frac{1}{2}\)) + (\(\frac{27}{23}+\frac{16}{23}\)) - \(\frac{5}{27}\)
=\(\frac{109}{10}+\frac{43}{23}-\frac{5}{27}\)
=12,58438003
hay \(\approx12,58\)