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:
\(2ab+2bc+2ca=-1\)

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

\(x^3+2x-3=0\)

\(\Leftrightarrow x^3-x+3x-3=0\)

\(\Leftrightarrow x\left(x^2-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+3\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2+x+\dfrac{1}{4}\right)+3-\dfrac{1}{4}\right]=0\) 

\(\Leftrightarrow\left(x-1\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]=0\) (*)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\) (**)

Từ (*);(**) => \(x-1=0\) =>x=1

Vậy x=1