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x = 1/8 - y/4 = (1-2y)/8
<=> x = 5*8/(1-2y) ; thấy 1-2y là số lẻ nên UCLN(8,1-2y) = 1
do đó x/8 = 5/(1-2y) (*)
x, y nguyên khi 1-2y phải là ước của 5
* 1-2y = -1 => y = 1 => x = -40
* 1-2y = 1 => y = 0 => x = 40
* 1-2y = -5 => y = 3 => x = -8
* 1-2y = 5 => y = -2 => x = 8
vậy có 4 cặp (x,y) nguyên (-40,1) ; (40, 0) ; (-8, -5) ; (8, 5) .
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
Vậy S = { 0, 6}
![](https://rs.olm.vn/images/avt/0.png?1311)
(8 - 5x) (x + 2) + 4(x - 2) (x + 1) + 2(x - 2) (x + 2) = 0
=> (x + 2) [ (8 - 5x) + 4(x + 1) + 2(x - 2)] = 0
=> (x + 2) (8 - 5x + 4x + 4 + 2x - 4) = 0
=> (x + 2) (x + 8) = 0
=> x + 2 = 0 hoặc x + 8 = 0
=> x = -2 hoặc x = -8
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : (x - 1)(x + 2) = x2 - 8
=> x2 + x - 2 = x2 - 8
=> x - 2 = -8
=> x = -6
Vậy x = -6
( x - 1 )( x + 2 ) = x2 - 8
⇔ x2 + x - 2 - x2 + 8 = 0
⇔ x + 6 = 0
⇔ x = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-x\left(x+2\right)=6\)
\(\Leftrightarrow x^2-x^2-2x=6\)
<=> -2x = 6
<=> x = -3
\(3x\left(x-2\right)+2x\left(2-x\right)=x^2-8\)
\(\Leftrightarrow3x\left(x-2\right)-2x\left(x-2\right)=x^2-8\)
\(\Leftrightarrow\left(x-2\right)\left(3x-2x\right)=x^2-8\)
\(\Leftrightarrow\left(x-2\right)x=x^2-8\)
\(\Leftrightarrow x^2-2x=x^2-8\)
\(\Leftrightarrow2x=8\)
<=> x = 4
a/ \(x^2-x\left(x+2\right)=6\)
<=> \(x^2-x^2-2x=6\)
<=> \(-2x=6\)
<=> \(x=-3\)
b/ \(3x\left(x-2\right)+2x\left(2-x\right)=x^2-8\)
<=> \(3x^2-6x+4x-2x^2=x^2-8\)
<=> \(3x^2-2x-2x^2-x^2+8=0\)
<=> \(-2x+8=0\)
<=> \(-2x=-8\)
<=> \(x=4\)
c/ \(3\left(5x-1\right)-x\left(x+1\right)+x^2=14\)
<=> \(15x-3-x^2-x+x^2=14\)
<=> \(14x-3=14\)
<=> \(-3=14-14x\)
<=> \(14\left(1-x\right)=-3\)
<=> \(1-x=\frac{-3}{14}\)
<=> \(-x=\frac{-3}{14}-1\)
<=> \(x=\frac{3}{14}+1\)
<=> \(x=\frac{17}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x^2-2x+1-x^2+4=8\\ \Leftrightarrow-2x=8-4-1\\ \Leftrightarrow-2x=3\\ \Leftrightarrow x=-\dfrac{3}{2}\)
\(\Leftrightarrow x^2-2x+1-x^2+4=8\\ \Leftrightarrow-2x=3\Leftrightarrow x=-\dfrac{3}{2}\)