Lời giải :

\(x^2-2014xy-2016xz+\left(2015^2-1\right)yz\)

\(=x^2-2014xy-2016xz+\left(2015-1\right)\left(2015+1\right)yz\)

\(=x^2-2014xy-2016xz+2014\cdot2016\cdot yz\)

\(=x\left(x-2014y\right)-2016z\left(x-2014y\right)\)

\(=\left(x-2014y\right)\left(x-2016z\right)\)

x² - 2014xy - 2016xz + (2015² - 1)yz

= x² - 2014xy - 2016xz + (2015 - 1)(2015 + 1)yz

= x² - 2014xy - 2016xz + 2014.2016.yz

= (x² - 2014xy) - (2016xz - 2014.2016.yz)

= x(x - 2014y) - 2016z(x - 2014y)

= (x - 2014y)(x - 2016z)

#TT

1/ \(\left(a-b\right)\left(a^2+3ab+b^2\right)+\left(a+b\right)^3+ab\left(b-a\right)=\left(a^2+2ab+b^2+ab\right)\left(a-b\right)+\left(a+b\right)^3+ab\left(b-a\right)\)= \(\left(a^2+2ab+b^2\right)\left(a-b\right)+\left(a+b\right)ab+\left(a-b\right)^3-ab\left(a-b\right)\)

= \(\left(a+b\right)^2\left(a-b\right)+\left(a+b\right)^3\)

= \(\left(a+b\right)^2\left(a-b+a+b\right)=2a\left(a+b\right)^2\)

k mình nhé!

Giup mình với nka^^

Có: \(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\Leftrightarrow\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\)\(\Leftrightarrow x=y=z\)

Lại có: \(x^{2015}+y^{2015}+z^{2015}=3^{2016}\)

\(\Leftrightarrow x^{2015}+x^{2015}+x^{2015}=3^{2016}\)

\(\Leftrightarrow3x^{2015}=3^{2016}\)

\(\Leftrightarrow x=3\)

Vậy \(x=y=z=3\)

ai đó làm giúp mình , mình tích cho

nhân 2 vế cho 2

=>2x²+2y²+2z²=2xy+2yz+2zx

=>2x²+2y²+2z²-2xy-2yz-2zx=0

=>(2x²-2xy)+(2y²-2yz)+(2z²-2zx)=0

=>(x-y)²+(y-z)²+(z-x)²=0

mà (x-y)² >= 0 với mọi x,y

(y-z)² >= 0 với mọi y,z

(z-x)² >=0 với mọi z,x

=>(x-y)²+(y-z)²+(z-x)² >= 0

mà theo đề:(x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

=>x=y

   y=z

   z=x

hay x=y=z

do đó x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=3²⁰¹⁶

<=>x²⁰¹⁵+x²⁰¹⁵+x²⁰¹⁵=3²⁰¹⁶

<=>3x²⁰¹⁵=3²⁰¹⁶<=>x²⁰¹⁵=3²⁰¹⁶:3=3²⁰¹⁵<=>x=2015

Vậy x=y=z=2015

\(x^2+y^2+z^2=xy+yz+xz\)

\(2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Rightarrow}}x=y=z\)

\(\Rightarrow x^{2015}+y^{2015}+z^{2015}=x^{2015}+x^{2015}+x^{2015}=3x^{2015}\)

\(\Rightarrow3x^{2015}=3^{2016}\)

\(\Rightarrow x^{2015}=3^{2015}\)

\(\Rightarrow x=3\)

Vậy \(x=y=z=3\)