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\(\frac{27.18+27.103-120.27}{15.33+33.12}=\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}=\frac{27.1}{33.27}=\frac{1.1}{33.1}=\frac{1}{33}\)
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\(\frac{27\cdot18+27\cdot103-27\cdot120}{15\cdot33+12\cdot33}=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)\(=\frac{27\cdot1}{33\cdot27}=\frac{1}{33}\)
\(\Leftrightarrow x-\frac{1}{3}=\left(+-\right)\frac{5}{6}\)
nếu \(x-\frac{1}{3}=\frac{5}{6}\) nếu \(x-\frac{1}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}+\frac{1}{3}\) \(\Leftrightarrow x=-\frac{5}{6}+\frac{1}{3}\)
\(\Leftrightarrow x=\frac{7}{6}\) \(\Leftrightarrow x=-\frac{1}{2}\)
Vậy ....
nhớ k mk nha bạn , mk nhanh nhất
thanks
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\)\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)
\(=\)\(\frac{27.1}{33.27}\)
\(=\)\(\frac{1}{33}\)
1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
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)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)
\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)
mik dang ban moi giai duoc mot bai ha, sorry
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
