D

\(\dfrac{x}{7}=\dfrac{x-12}{21}\)

\(21x=7x-84\\ 21x-7x=-84\\ 14x=-84\\ x=-6\)

1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)

\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)

2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)

\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)

3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)

\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)

\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)

\(\Leftrightarrow\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{14}-...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{5}-\dfrac{98}{515}\\ \Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\\ \Leftrightarrow y+3=103\\ \Leftrightarrow y=100\)

Vậy...........................................

\(\dfrac{2011+2012+2013}{2012+2013+2014}=\dfrac{6036}{6039}=\dfrac{2012}{2013}\)

Có : \(2009+2010>\dfrac{2009}{2010}\) ; \(2011+2012>\dfrac{2011}{2012}\)

\(\dfrac{2011}{2010}>1\) ; \(\dfrac{2010}{2011}< 1\) \(\Rightarrow\dfrac{2011}{2010}>\dfrac{2010}{2011}\)

Ta có : \(2009+2010+\dfrac{2011}{2010}+2011+2012>\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2011}{2012}\)

\(\Leftrightarrow B>A\)

Hay \(A< B\)

Ta có : \(\dfrac{-5}{14}=\dfrac{-20}{56}\) ; \(\dfrac{-4}{11}=\dfrac{-20}{55}\)

Vì \(\dfrac{-20}{56}>\dfrac{-20}{55}\) nên \(\dfrac{-5}{14}>\dfrac{-4}{11}\)

Chiều cao hình bình hành là:

36 x 2/3 = 24(cm)

Diện tích hình bình hành là:

36 x 24 = 864(cm2)

           Đáp số: 864cm2

bn chỉ cần vẽ hình tròn tâm O bán kính 4cm thì M sẽ thuộc bất cứ điểm nào trên đường tròn

là 4 vì \(S^2=\dfrac{3^2+2^2+1^2+0^2+1^2+2^2+3^2}{7}=4\)