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

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

\(\frac{27\cdot18+27\cdot103-27\cdot120}{15\cdot33+12\cdot33}=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)\(=\frac{27\cdot1}{33\cdot27}=\frac{1}{33}\)

\(\Leftrightarrow x-\frac{1}{3}=\left(+-\right)\frac{5}{6}\)

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

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

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

\(=\)\(\frac{27.1}{33.27}\)

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

\(|x-\frac{1}{3}|=\frac{5}{6}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x-\frac{1}{3}=\frac{5}{6}\\x-\frac{1}{3}=-\frac{5}{6}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{5}{6}+\frac{1}{3}\\x=-\frac{5}{6}+\frac{1}{3}\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{7}{6}\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{7}{6};-\frac{1}{2}\)

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)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)

\(\frac{298}{719}:\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}:0-\frac{2011}{2012}\)

Giá trị phép tính không tồn tại

\(\frac{298}{719}\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}\cdot0-\frac{2011}{2012}\)

\(=0-\frac{2011}{2012}=-\frac{2011}{2012}\)

\(\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\div\frac{7}{12}\)

\(=\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\times\frac{12}{7}\)

\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-4}{9}+\frac{5}{6}\right)\)

\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-8}{18}+\frac{15}{18}\right)\)

\(=\frac{2}{3}+\frac{4}{7}.\frac{-7}{18}\)

\(=\frac{6}{9}+\frac{-2}{9}\)

\(=\frac{4}{9}\)

\(\frac{298}{719}\div\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}\div\left(\frac{3}{12}+\frac{1}{12}-\frac{4}{12}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}\div0-\frac{2011}{2012}\)

Vậy biểu thức đại số này không xác định

4/7 ở đâu ra vậy

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

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

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

= 1/33

ai k mk thì mk k lại