Bài 1 :
a, \(xy+3x-2y-6=2\Leftrightarrow y\left(x-2\right)+3\left(x-2\right)=2\)
<=> \(\left(x-2\right)\left(y+3\right)=2\)
Vì x,y nguyên nêu x-2 , y+3 cũng nguyên =>(x-2) và (y+3 )
Ta có bảng .............
1.
a) Ta có: yx + 3x - 2y - 6
= x (y + 3) - 2 (y + 3)
= (x - 2)(y + 3) = 2
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1;y+3=2\\x-2=-1;y+3=-2\\x-2=2;y+3=1\\x-2=-2;y+3=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3;y=-1\\x=1;y=-5\\x=4;y=-2\\x=0;y=-4\end{matrix}\right.\)
Vậy (x,y) \(\in\) {(3;-1);(1;-5);(4;-2);(0;-4)}.
b) Ta có: xy - x + 5y = 7
\(\Leftrightarrow\) xy - x + 5y - 5 = 2
\(\Leftrightarrow\) x (y - 1) + 5 (y - 1) = 2
\(\Leftrightarrow\) (x + 5)(y - 1) = 2
...(Làm tương tự câu a)...
Vậy (x,y) \(\in\) {(-4;3);(-6;-1);(-3;2);(-7;0)}.
2.
a) Ta có: (x2 + x + 1)(x2 + x + 2) = 12 (*)
\(x^2+x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \dfrac{3}{4}>0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\Rightarrow x^2+x+1>0\forall x\)
Do đó x2 + x + 1 và x2 + x + 2 không thể là số âm.
Từ (*) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x^2+x+1=3;x^2+x+2=4\\x^2+x+1=4;x^2+x+2=3\left(l\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x+1=2;x^2+x+2=6\left(l\right)\\x^2+x+1=6;x^2+x+2=2\left(l\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x+1=1;x^2+x+2=12\left(l\right)\\x^2+x+1=12;x^2+x+2=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x^2+x+1-3=0\Leftrightarrow x^2+x-2=0\)
Ta có: x2 + x - 2
= x2 + 2x - x - 2
= x (x + 2) - (x + 2)
= (x - 1)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy x1 = 1; x2 = -2.
b) Đặt \(x^2+4x+8=t\)
Ta có: \(t^2+3xt+2x^2\)
\(=t^2+2xt+xt+2x^2\\ =t\left(t+2x\right)+x\left(t+2x\right)\\ =\left(t+x\right)\left(t+2x\right)\)
Vậy \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+x\right)\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\).
c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+5=t\)
Ta có: \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\\ =t^2-25=\left(t-5\right)\left(t+5\right)\)
Vậy ...
