\(A=-2\)

\(\Leftrightarrow5x^2+y^2+4xy-6x-2y=-2\)

\(\Leftrightarrow4x^2+x^2+y^2+4xy-4x-2x-2y+1+1=0\)

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

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

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

Mà \(\left(2x+y-1\right)^2+\left(x-1\right)^2\ge0\)nên (1) xảy ra

\(\Leftrightarrow\hept{\begin{cases}2x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-1\\x=1\end{cases}}\)

\(\Rightarrow B=1^{2015}.\left(-1\right)^{2016}-1^{2016}.\left(-1\right)^{2017}+2014\)

\(=1+1+2014=2016\)

Giúp mình với đang cần gấp!!

Ta có: A = -2

=> 5x² + y² + 4xy - 6x - 2y = -2

=> 5x² + y² + 4xy - 6x - 2y + 2 = 0

=> (4x² + 4xy + y²) - 2(2x + y) + 1 + (x² - 2x + 1) = 0

=> (2x + y)² - 2(2x + y) + 1 + (x - 1)² = 0

=> (2x + y - 1)² + (x - 1)² = 0

       <=> \(\hept{\begin{cases}2x+y-1=0\\x-1=0\end{cases}}\)

        <=> \(\hept{\begin{cases}y=1-2x\\x=1\end{cases}}\)

        <=> \(\hept{\begin{cases}y=1-2.1=-1\\x=1\end{cases}}\)

Với x = 1; y = -1 => B = 1²⁰¹⁵.(-1)²⁰¹⁶ - 1²⁰¹⁶.(-1)²⁰¹⁷ + 2014

                                    = 1 + 1 + 2014 = 2016

$\textbf{1.}$

Ta có $5^{2017}+5^{2015}=5^{2015}(5^2+1)$

$=5^{2015}\cdot26$

$=5^{2015}\cdot13\cdot2.$

Vì $13\mid13\cdot2$ nên $13\mid\left(5^{2017}+5^{2015}\right).$

Vậy $5^{2017}+5^{2015}$ chia hết cho $13.$

$\textbf{2.}$

Giả sử $a^{2014}+b^{2015}+c^{2016}\vdots6.$

Ta có $a^{2016}-a^{2014}=a^{2014}(a^2-1)$

$=a^{2014}(a-1)(a+1).$

Vì $a(a-1)(a+1)\vdots6$ nên $a^{2014}(a-1)(a+1)=a^{2013}\cdot a(a-1)(a+1)\vdots6.$

Suy ra $a^{2016}\equiv a^{2014}\pmod6.$

Tương tự, $b^{2017}-b^{2015}=b^{2015}(b^2-1)$

$=b^{2014}\cdot b(b-1)(b+1)\vdots6,$ nên $b^{2017}\equiv b^{2015}\pmod6.$

Lại có $c^{2018}-c^{2016}=c^{2016}(c^2-1)$

$=c^{2015}\cdot c(c-1)(c+1)\vdots6,$ nên $c^{2018}\equiv c^{2016}\pmod6.$

Cộng ba đồng dư trên, $a^{2016}+b^{2017}+c^{2018}\equiva^{2014}+b^{2015}+c^{2016}\equiv0\pmod6.$

Vậy $a^{2016}+b^{2017}+c^{2018}$ chia hết cho $6.$

Ta có 

a²+b²+c² = ab+bc+ca

<=> 2(a²+b²+c²)= 2(ab+bc+ca)

<=> (a - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> a = b = c

Thế vào pt thứ (2) ta được

a⁸ + b⁸ + c⁸ = 3

<=> 3a⁸ = 3

<=> a⁸ = 1

<=> a = b = c = 1(3) hoặc a = b = c = - 1(4)

Từ (3) => P = 1 + 1 - 1 = 1

Từ (4) => P = - 1 + 1 + 1 = 1