\(42:x+36:x=24\)

\(\Leftrightarrow\left(42+36\right):x=24\)

\(\Leftrightarrow78:x=24\)

\(\Leftrightarrow x=78:24=\frac{13}{4}\)

\(1+2+\left(-3\right)+\left(-4\right)+....+2013+2014+\left(-2015\right)+\left(-2016\right)\)

\(=\left(1-3\right)+\left(2-4\right)+...+\left(2013-2015\right)+\left(2014-2016\right)\)

\(=\left(-2\right)+\left(-2\right)+.....+\left(-2\right)+\left(-2\right)\)

\(=\left(-2\right)\times1008=-2016\)

  1+2+(-3)+(-4)+...+2013+2014+(-2015)+(-2016)

= (1-3)+(2-4)+...+(2013-2015)+(2014-2016)

= (-2)+(-2)+...+(-2)+(-2)

= (-2) x 1008

= -2016

cảm ơn bạn nha :))

\(\hept{\begin{cases}x^2+y^2+xy+1=2x_{ }\left(1\right)\\x\left(x+y\right)^2+x-2=2y^2_{ }_{ }\left(2\right)\end{cases}}\)

lấy 2*(1)+(2) ta có

\(2x^2+2xy+x\left(x+y\right)^2+x=4x\Leftrightarrow\orbr{\begin{cases}x=0\\2\left(x+y\right)+\left(x+y\right)^2=3\end{cases}}\)

\(x=0\Rightarrow\left(1\right)\Leftrightarrow y^2+1=0\)vô nghiệm

\(2\left(x+y\right)+\left(x+y\right)^2=3\Leftrightarrow\orbr{\begin{cases}x+y=1\\x+y=-3\end{cases}}\)

thế \(y=1-x\) vào (2) ta có \(x+x-2=2\left(1-x\right)^2\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=0\\x=2\Rightarrow y=-1\end{cases}}\)

thế \(y=-3-x\) vào (2) ta có \(9x+x-2=2\left(-3-x\right)^2\Leftrightarrow x^2+x+10=0\) vô nghiệm

vậy hệ cso hai nghiệm

\(M=3\left|x-2013\right|+\left|x+1\right|+\left|x+2014\right|\)

hay \(M=\left|2013-x\right|+\left|x+1\right|+\left|2013-x\right|+\left|x+2014\right|+\left|2013-x\right|\)

mà ta có 

\(\hept{\begin{cases}\left|2013-x\right|\ge0\\\left|2013-x\right|+\left|x+1\right|\ge\left|2013-x+x+1\right|=2014\\\left|2013-x\right|+\left|x+2014\right|\ge\left|2013-x+x+2014\right|=4027\end{cases}}\)

Vậy \(M\ge2014+4027=6041\)

dấu bằng xay ra khi x=2013