Ta có \(x^{2014}+x^{2012}+1=x^{2014}-x+x^{2012}-x^2+x^2+x+1\)

=\(x\left(x^{2013}-1\right)+x^2\left(x^{2010}-1\right)+x^2+x+1=x\left(x^3-1\right)\left(...\right)+x^2\left(x^3-1\right)\left(...\right)+x^2+x+1\)

=\(\left(x^2+x+1\right)\left(...\right)\RightarrowĐPCM\)

x = 16 nhé

x^2+x*y-2012*x-2013*y-2014=0

<=> x^2+x*y+x-2013*x-2013*y-2014=0

<=>x*(x+y+1)-2013*(x+y+1)=0

<=> (x-2013)*(x+y+1)=1

do x,y nguyen nen

(x,y) la (2014;-2014);(2012;-2014)

Nguyễn Việt Lâm

Chỉ em câu này với ạ

\(\left(x^2-2x+3\right)\left(\frac{1}{2x}-5\right)\)

\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)

\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)

\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)

\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)

học tốt

(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15

\(x^{n-1}\left(x+y\right)-y\left(x^{n-1}+y^{n-1}\right)\)

\(=x^{n-1}.x+x^{n-1}.y-y.x^{n-1}-y.y^{n-1}\)

\(=x^{n-1+1}+x^{n-1}y-x^{n-1}y-y^{n-1+1}=x^n-y^n\)

\(\left(x-4\right)^2=\left(2x+1\right)^2\)

\(\Leftrightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\3\left(x-1\right)=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)

(x-4)² = (2x+1)²

=> x-4 = 2x +1

    x - 2x = 1 +4

   -x = 5

   x=-5