N=\(\frac{27.18+27.103-120.27}{15.33+33.12}\)

N=\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)

N=\(\frac{21.1}{33.27}\)

N=\(\frac{1}{33}\)

Vậy N=\(\frac{1}{33}\)

N=1/33

\(\frac{27.18+27.103-120.27}{15.33+33.12}=\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}=\frac{27.1}{33.27}=\frac{1.1}{33.1}=\frac{1}{33}\)

Đúng 100%

Good Luck

^.^

\(=\frac{1}{33}\)

= 1/33

ai k mk thì mk k lại

1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)

1) \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\) \(=\dfrac{-5}{2}:\dfrac{1}{4}=-10\)

\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)

mik dang ban moi giai duoc mot bai ha, sorry

k biet

\(\frac{27\cdot18\cdot27\cdot103-102\cdot27}{15\cdot33+33\cdot12}\)

=\(\frac{27\cdot\left(18+103-102\right)}{33\cdot\left(15+12\right)}\)

=\(\frac{27\cdot19}{33\cdot27}\)

=\(\frac{19}{33}\)

a) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{3}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3}{12}+\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3+1+4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\dfrac{2}{3}-\dfrac{2011}{2012}=\dfrac{298}{719}\cdot\dfrac{3}{2}-\dfrac{2011}{2012}=\dfrac{149.3}{719.1}-\dfrac{2011}{2012}=\dfrac{447}{719}-\dfrac{2011}{2012}=\dfrac{889364}{1446628}-\dfrac{1445909}{1446628}=\dfrac{889364-1445909}{1446628}=-\dfrac{556545}{1446628}.\)b)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)

\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)