(x-2015)^x+1 - (x-2015)^x+2015

=>x-2015= 0;1;-1

x-2015=0 =>x=2015

x-2015=1 =>x=2016

x-2015=-1 =>x=2014

\(x\left(x-2015\right)+\left(x-2015\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-2015\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\x-2015=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=2015\end{cases}}\)

\(x\left(x-2015\right)+x-2015=0\)

\(x\left(x-2015\right)+\left(x-2015\right)=\left(x-2015\right)\left(x+1\right)=0\)

TH1 :\(x+1=0\)

\(x=-1\)

TH2 : \(x-2015=0\)

\(x=2015\)

\(b,\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow x-2015>0\Leftrightarrow x>2015\\ d,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x>0\\x+6>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x< 0\\x+6< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-6< x< 3\)

x=5

b: \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=-4\end{matrix}\right.\)

a/ => | x - 7 | = 5 + x

TH1: x - 7 = 5 + x => 0x = 12 (VN)

TH2: x - 7 = - 5 - x => 2x = 2 => x = 1

Vậy x = 1

b/ \(\Rightarrow\int^{x-2015\ge0}_{x+2015<0}\)  \(\Rightarrow\int^{x\ge2015}_{x<-2015}\) (vô lí)

hoặc \(\int^{x-2015<0}_{x+2015\ge0}\)   \(\Rightarrow\int^{x<2015}_{x\ge-2015}\) \(\Rightarrow-2015\le x<2015\)

Ta có: \(8x\left(x-2015\right)-x+2015=0\)

\(\Rightarrow8x\left(x-2015\right)-\left(x-2015\right)=0\)

\(\Rightarrow\left(8x-1\right)\left(x-2015\right)=0\)

\(\Rightarrow\orbr{\begin{cases}8x-1=0\\x-2015=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{8}\\x=2015\end{cases}}}\)

Vậy \(x=\left\{\frac{1}{8};2015\right\}\)

       8x(x - 2015) - (x - 2015) = 0

<=> (8x - 1)(x - 2015) = 0

<=> ...........  đến đây thì dễ rồi :)))) 

8x(x-2015)-x+2015=0

8x(x-2015)-(x-2015)=0

(8x-1)(x-2015)=0

\(\orbr{\begin{cases}8x-1=0\\x-2015=0\end{cases}}\)=>\(\orbr{\begin{cases}8x=-1\\x=2015\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{1}{8}\\x=2015\end{cases}}\)

vậy x \(\in\){\(\frac{1}{8}\);2015}

a. (x-1).(x-5)>0

Suy ra (x-1).(x-5) la so nguyen duong

Ta co : so duong = so duong . so duong = so am. so am 

Suy ra (x-5)nho nhat = -5 vay x = 0  Suy ra x = {0;1;2;3;4;5;6;.......................................}

Ma (x-1).0 hoac (x-5). 0=0

Suy ra 1 , 5 ko thuoc x

Suy ra x = {0;2;3;4;6;.........................................}

tick cho to roi to lam tiep phan b

C1: 

\(1-x>0\Leftrightarrow x< 1\)

C2:

\(2k+1< 0\)

\(\Leftrightarrow k< \dfrac{-1}{2}\)

C3:

\(x+1\ne0\Leftrightarrow x\ne1\)

1B

2B

3B

4D

Ta có / x + 2013 / + / x + 2014 / + / x + 2015 / = 0x

\(=>x+2013+x+2014+x+2015=0\)

\(=>\hept{\begin{cases}x+2013=0\\x+2014=0\\x+2015=0\end{cases}=>\hept{\begin{cases}x=-2013\\x=-2014\\x=-21015\end{cases}}}\)