<=>(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

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

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

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

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

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

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

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

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

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

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

thanks

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)

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