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

\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)

ĐK:\(x\ge\frac{2015}{2016}\)

\(\Leftrightarrow2015\left(\sqrt{2015x-2014}-1\right)+\sqrt{2016x-2015}-1=0\)

\(\Leftrightarrow2015\frac{2015x-2014-1}{\sqrt{2015x-2014}+1}+\frac{2016x-2015-1}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015x-2015}{\sqrt{2015x-2014}+1}+\frac{2016x-2016}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015\left(x-1\right)}{\sqrt{2015x-2014}+1}+\frac{2016\left(x-1\right)}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}\right)=0\)

Dễ thấy: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

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.

Theo đề bài ta có

\(f\left(x\right)=x^{2017}-2016.x^{2016}+2016.x^{2015}-...+2016.x-1\)

Với \(f\left(2015\right)\)thì \(x=2015,x+1=2016\)

\(\Rightarrow f\left(x\right)=x^{2017}-\left(x+1\right).x^{2016}+\left(x+1\right).x^{2015}-...+\left(x+1\right).x-1\)

\(\Rightarrow f\left(x\right)=x^{2017}-x^{2017}-x^{2016}+x^{2016}+x^{2015}-...+x^2+x-1\)

\(\Rightarrow f\left(x\right)=x-1\)

\(\Rightarrow f\left(2015\right)=2015-1=2014\)

Vậy f(2015)=2014

\(x\ge-3\)

\(x^4\left(\sqrt{x+3}-2\right)+2016\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{x^4\left(x-1\right)}{\sqrt{x+3}+2}+2016\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x^4}{\sqrt{x+3}+2}+2016\right)=0\)

\(\Leftrightarrow x-1=0\) (do \(\dfrac{x^4}{\sqrt{x+3}+2}+2016>0\) \(\forall x\ge-3\) )

\(\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)