-5/ 12. (2/11+ 9/11+ (-1)
=-5/ 12. [(11/11+(-1)]

=-5/ 12. (1-1)

= -5/12. 0

=0

\(=\dfrac{-5}{12}\times\left(\dfrac{2}{11}+\dfrac{9}{11}-1\right)\)

\(=\dfrac{-5}{12}\times0=0\)

=−512×0=0

\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)

\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)

\(D=12\cdot\frac{49}{303}\)

\(D=\frac{588}{303}\)

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)

=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)

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

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

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

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\)

=\(\left(4-\dfrac{5}{12}\right).\dfrac{1}{2}+\dfrac{5}{24}\)

=\(4.\dfrac{1}{2}-\dfrac{5}{12}.\dfrac{1}{2}+\dfrac{5}{24}\)

=\(2-\dfrac{5}{24}+\dfrac{5}{24}=2\)

Giải:

\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\) 

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

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

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

\(=\dfrac{1}{3}.\left(-1\right)\) 

\(=\dfrac{-1}{3}\) 

\(\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\) 

\(=\dfrac{43}{12}:2+\dfrac{5}{24}\) 

\(=\dfrac{43}{24}+\dfrac{5}{24}\) 

\(=2\)

f: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=\dfrac{-5}{19}\)

i: \(=\left(\dfrac{9}{24}-\dfrac{18}{24}+\dfrac{14}{24}\right)\cdot\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{5}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}\)

=1/4+1/2=3/4

` 7/19 . 8/11 + 3/11 . 7/19 + (-12)/19 `

`= 7/19 . ( 8/11 + 3/11 ) + (-12)/19 `

`= 7/19 . 11/11 + (-12)/19`

`= 7/19 . 1 + (-12)/19 `

`= 7/19 + (-12)/19 `

`= -5/19 `

`( 3/8 + (-3)/4 + 7/12 ) : 5/6 + 1/2`

`= 3/8 + (-3)4 + 7/12 . 6/5 + 1/2`

`= ( 9+(-18) + 14)/24 . 6/5 + 1/2`

`= 5/24 . 6/5 + 1/2`

`= 1/4 + 1/2 `

`= 3/4`

\(a.\)

\(\dfrac{27}{13}-\dfrac{106}{111}+-\dfrac{5}{111}=\dfrac{27}{13}-\dfrac{106}{111}-\dfrac{5}{111}=\dfrac{27}{13}-\left(\dfrac{106+6}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)

\(b.\)

\(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)

\(c.\)

\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+-\dfrac{6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{25}{31}+\dfrac{6}{31}\right)=1-1=0\)

a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}=\dfrac{27}{13}+\left(\dfrac{-106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}+-1=\dfrac{14}{13}\) 

b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\left(\dfrac{7}{19}+\dfrac{12}{19}\right)=\dfrac{12}{11}+1=\dfrac{23}{11}\) 

c)\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{-25}{31}+\dfrac{-6}{31}\right)=1+-1=0\)

\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)

\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)

\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)

\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)

\(\dfrac{-1}{6}\)

\(\dfrac{-1}{6}\)

\(\dfrac{-11}{15}\)

\(\dfrac{-9}{8}\)

\(d,-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

\(e,\dfrac{5}{7}.\dfrac{25}{2}-\dfrac{5}{7}.\dfrac{11}{2}=\dfrac{5}{7}.\left(\dfrac{25}{2}-\dfrac{11}{2}\right)=\dfrac{5}{7}.7=5\)

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)