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14 tháng 2 2020

We have two cases:

+) If \(x\ge2016\)then \(x-2016\ge0\Rightarrow\left|x-2016\right|=x-2016\)

Equation becomes: \(x-2016=2016x\)

\(\Leftrightarrow2015x=-2016\Leftrightarrow x=\frac{-2016}{2015}\)(not satisfied)

+) If \(x< 2016\)then \(x-2016< 0\Rightarrow\left|x-2016\right|=2016-x\)

Equation becomes: \(2016-x=2016x\)

\(\Leftrightarrow2017x=2016\Leftrightarrow x=\frac{2016}{2017}\)(satisfied)

So ​\(x=\frac{2016}{2017}\)

14 tháng 2 2020

ĐK: x>0.

Pt\(\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2015}{2016}\left(l\right)\\x=\frac{2016}{2017}\end{matrix}\right.\)

Vậy \(S=\left\{\frac{2016}{2017}\right\}\)

Ta có: \(\left|x-2016\right|=2016x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2016-2016x=0\\x-2016+2016x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2015x=2016\\2017x=2016\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2016}{2015}\\x=\frac{2016}{2017}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{2016}{2015};\frac{2016}{2017}\right\}\)

18 tháng 9 2015

x=2015

=> x+1=2016

=> A=x2016-(x+1).x2015+(x+1).x2014-(x+1).x2013+...+(x+1)x2-(x+1)x+2016

=x2016-x2016-x2015+x2015+x2014-x2014-x2013+...+x3+x2-x2-x+2016

=-x+2016

=-2015+2016

=1

Vậy A=1.

8 tháng 1 2017

ko bít làm à

8 tháng 1 2017

k bik nên mới hỏi

11 tháng 1 2017

Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)

20 tháng 10 2019

\(x^4+2016x^2+2017x+2016\)

\(=x^4+2016x^2+2016x+x+2016\)

\(=\left(x^4+x\right)+\left(2016x^2+2016x+2016\right)\)

\(=x\left(x^3+1\right)+2016\left(x^2+x+1\right)\)

\(=x\left(x+1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+2016\right)\)