Chính là 0

\(|2015x-2014|=|2015x+2014|\)

\(\Leftrightarrow\orbr{\begin{cases}-2015x+2014=|2015x+2014|\left(l\right)\\2015x-2014=|2015x+2014|\left(n\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2015x+2014=-2015x+2014\\2015x+2014=2015x-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}4030x=0\\0x=-4028\left(l\right)\end{cases}\Leftrightarrow}4030x=0\Leftrightarrow x=0}\)

<=>(x²-x)-(2015x-2014)=0

<=>x(x-1)-2014(x-1)=0

<=>(x-2014)(x-1)

<=>x-2014=0

hoặc x-1=0

<=>x=2014

hoặc x=1

h

\(x^2-2015x+2014=0\)

\(\Leftrightarrow x^2-2014x-x+2014=0\)

\(\Leftrightarrow x\left(x-2014\right)-\left(x-2014\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2014\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2014=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2014\end{cases}}\)

\(x^2-2015x+2014=0\)

\(x^2-x-2014x+2014=0\)

\(x\left(x-1\right)-2014\left(x-1\right)=0\)

\(\left(x-1\right)\left(x-2014\right)=0\)

TH1:x -1 = 0

=>x=1

TH2 : x-2014=0

=> x=2014

\(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(x\left(x-4\right)\left(x+4\right)=0\)

TH1: x=0

TH2:x-4=0

=> x= 4

TH3: x+4=0

=> x=(-4)

Hok tốt

a) \(P\left(x\right)=0\Rightarrow x^{2016}-x^{2014}=0\Rightarrow x^{2014}\left(x^2-1\right)=0\)

TH1: \(x^{2014}=0\Rightarrow x=0\)

TH2: \(x^2-1=0\Rightarrow x=\pm1\)

Vậy \(P\left(x\right)\) có nghiệm là \(x=0,x=1,x=-1\)

b) Xét \(x< 0\)

Ta có: \(x^{2016}>0\Rightarrow-x^{2016}< 0\); \(2015x< 0\)

\(\Rightarrow Q\left(x\right)=-x^{2016}+2015x-1< 0\)

Vậy \(Q\left(x\right)\) không có nghiệm âm

a, Đặt \(P\left(x\right)=x^{2016}-x^{2014}=0\Leftrightarrow x^{2014}\left(x^2-1\right)=0\Leftrightarrow x=0;x=-1;x=1\)

=> x là số nguyên dượng

=> x+1+x+2+x+3+x+4+...+x+2014=|x+1|+|x+2|+|x+3|+|x+4|+...+|x+2014|=2015

=> 2014x + (1+2+3+4+...+2014)=2015x

=> 1+2+3+4+...+2014=x

x^4-2014x^2+2015x-2014=0

<=>x⁴+x-2014x²+2014x-2014=0

<=>x.(x³+1)-2014.(x²-x+1)=0

<=>x.(x+1)(x²-x+1)-2014.(x²-x+1)=0

<=>(x²+x+1)[x.(x+1)-2014]=0

<=>x.(x+1)-2014=0 (vì x²+x+1 >0)

giải tiếp sao số xấu thế

|x+1| + |x+2| + |x+3| + .......... + |x+2014| = 2015x

Ta có :

|x+1| \(\ge\)0

|x+2| \(\ge\)0

|x+3| \(\ge\)0

..........

|x+2014| \(\ge\)0

=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| \(\ge\)0

=> 2015x \(\ge\)0

Mà 2015 \(\ge\)0

=> x \(\ge\)0 

=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| 

= x + 1 + x + 2 + x + 3 +.................... + x + 2014 = 2015x

=> 2014x + (1 + 2 + 3 +............ + 2014) = 2015x 

=> 1 + 2 + 3 + 4 + ........................ + 2014 = x 

=> x = 2029105