\(\left(x^2-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=\left(x^2\right)^2-2.x^2.2y+\left(2y\right)^2+\left(x^2-\left(\dfrac{1}{2}y\right)^2\right)\)

\(=x^4-4x^2y+4y^2+x^2-\dfrac{y^2}{4}\)

\(=x^4-4x^2y+x^2+\dfrac{15y^2}{4}\)

x² + 8 - xy + 6x - 5y = 0

<=> x² + 6x + 8 - y(x + 5) = 0

<=> x² + 6x + 8 = y(x + 5) 

<=> \(y=\dfrac{x^2+6x+8}{x+5}\)

Có \(y=\dfrac{x^2+6x+8}{x+5}=x+1+\dfrac{3}{x+5}\)

Để \(y\inℤ\Rightarrow3⋮x+5\Rightarrow x+5\inƯ\left(3\right)\)

\(\Rightarrow x+5\in\left\{1;3;-1;-3\right\}\Leftrightarrow x\in\left\{-4;-2;-6-8\right\}\)

Với x = -2 => y  = 0 

Với x = -4 => y = -1

Với x = - 6 => y = -8

Với x = -8 => y = -8

Vậy (x;y) = (-2 ; 0) ; (-4 ; -1) ; (-6 ; -8) ; (-8 ; -8) 

https://www.facebook.com/groups/1628547797542898

12+13+14+15+16+100+267548384543=

- Ta có: \(a^2+b^2+c^2=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=4\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2=4\left(ab+bc+ca\right)\)

- Vì \(4;\left(a+b+c\right)^2\) là các số chính phương

Nên \(ab+bc+ca\) phải là số chính phương (đpcm).

\(\left\{{}\begin{matrix}a+b+c=1\\a^2+b^2+c^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=1\\a^2+b^2+c^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ca=1\\a^2+b^2+c^2=2\end{matrix}\right.\)

trừ 2 vế ta được:
\(2ab+2bc+2ca=-1\)

\(\Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\)