\(\frac{6}{3}\cdot21+\frac{2}{7}\cdot11+\frac{9}{15}\cdot33+\frac{12}{15}\cdot114\)

\(< =>42+\frac{22}{7}+\frac{99}{5}+\frac{456}{5}\)

\(< =>\frac{316}{7}+111\)

\(< =>\frac{1093}{7}\)

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

\(\frac{27.18+27.103-27.102}{15.33+33.12}=\frac{27\left(18+103-102\right)}{33\left(15+12\right)}\)

\(=\frac{27.19}{33.27}=\frac{19}{33}\)

k nha

\(\frac{27.18+27.103-27.102}{15.33+33.12}=\frac{27.\left(18+103-102\right)}{33.\left(15+12\right)}=\frac{27.19}{33.27}=\frac{19}{33}\)

\(\frac{27.18+27.103-27.102}{15.33+33.12}=\frac{27.\left(18+103-102\right)}{33.\left(15+12\right)}\)

\(=\frac{513}{891}=\frac{19}{33}\)

a) 2(x+3)=-5(3-x)

<=>2x+6=-15+5x

<=>21=3x

<=>x=7

Vậy tập nghiệm T={7}

b)\(\dfrac{16.18-16.7}{15.33+33}=\dfrac{16.\left(18-7\right)}{33.\left(15+1\right)}=\dfrac{16.11}{3.11.16}=\dfrac{1}{3}\)

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

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

= 1/33

ai k mk thì mk k lại

Rút gọn: 

\(=\frac{16\left(18-4\right)}{33\left(15+1\right)}=\frac{16.14}{33.16}=\frac{14}{33}\)

So sánh thì bạn quy đồng lấy mẫu chung là 36

14/33

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

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

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 \(x=\frac{7}{6};-\frac{1}{2}\)