\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)

\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)

\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)

\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

\(\left(\dfrac{-2}{5}+\dfrac{3}{7}\right)-\left(\dfrac{4}{9}+\dfrac{12}{20}-\dfrac{13}{35}\right)+\dfrac{7}{35}\)

\(=-\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{4}{9}-\dfrac{3}{5}+\dfrac{13}{35}+\dfrac{7}{35}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{13}{35}+\dfrac{7}{35}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+1-\dfrac{4}{9}\\ =-\dfrac{4}{9}\)

2/5=12/30=40/100

2/5=12/30=40/100

Ta có:

\(\dfrac{4}{12}=\dfrac{1}{3}\)

\(\dfrac{5}{20}=\dfrac{1}{4}\)

\(\dfrac{21}{35}=\dfrac{3}{5}\)

\(\dfrac{17}{34}=\dfrac{1}{2}\)

Vậy \(\dfrac{17}{34}=\dfrac{1}{2}\)

không có phân số nào

\(\dfrac{17}{34}\)

\(\dfrac{-5}{12}+\dfrac{4}{37}+\dfrac{7}{12}-\dfrac{41}{37}+\dfrac{1}{2}\)

\(=\left(\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{1}{2}\right)+\left(\dfrac{4}{37}-\dfrac{41}{37}\right)\)

\(=\dfrac{2}{3}-1=-\dfrac{1}{3}\)