\(BC\left(68,85,170\right)< 2000=\left\{340,680,1020,1360,1700\right\}\)

câu hỏi tương tự cho nhanh

Ta có:

\(648=2^3.3^4\)

\(540=2^2.3^3.5\)

\(ƯCLN\left(648,540\right)=2^2.3^3=108\)

\(ƯC\left(648,540\right)=\left\{2,3,4,6,9,12,18,27,36,54,108\right\}\)

\(ƯCLN\left(648,540\right)=\left\{108\right\}\)

\(ƯC\left(648,540\right)=\left\{2,3,4,6,9,12,18,27,36,54,108\right\}\)

576+47+24

=(576+24)+47

=600+47

=647

\(160-\left(2^3.5^2-6.25\right)\)

\(=160-\left(8.25-6.25\right)\)

\(=160-\left[25.\left(8-6\right)\right]\)

\(=160-\left[25.2\right]\)

\(=160-50\)

\(=110\)

\(160-\left(2^3.5^2-6.25\right)\)

\(=160-\left(8.25-6.25\right)\)

\(=160-\left[25.\left(8-6\right)\right]\)

\(=160-\left[25.2\right]\)

\(=160-100\)

\(=60\)

\(\left|2x+1\right|=\dfrac{3}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1=\dfrac{3}{5}\\2x+1=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{5}-1\\2x=-\dfrac{3}{5}-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=-\dfrac{2}{5}\\2x=-\dfrac{8}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}:2\\x=-\dfrac{8}{5}:2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{5};-\dfrac{4}{5}\right\}\)

$aaaaaa:\left(3xa\right)=\frac{aaaaaa}{3a}=\frac{aaaaaa}{a}.\frac{1}{3}=111111:3=37037$