Question

1. Tìm x\(\varepsilon\) Z sao cho P/S sau là STN

a 26/x+3

b x+6/x+1

c 2x+1/x+3

2.tím các cặp số nguyên x,y biết

(x-1)(y+2)=-7

x(y-3)=-12

Answer 1

1. a, Để \(\dfrac{26}{x+3}\in N\Leftrightarrow26⋮x+3\)
=> x + 3 \(\inƯ\left(26\right)=\left\{\pm1;\pm2;\pm13;\pm26\right\}\)
=> x = -2; -4; -1; -5; 10; -16; 23; -29 (thỏa mãn)
b, Để \(\dfrac{x+6}{x+3}\in N\Leftrightarrow x+6⋮x+3\)
<=> x + 3 + 3 \(⋮x+3\)
<=> 3 \(⋮x+3\)
=> x + 3 \(\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
=> x = -2; -4; 0; -6 (thỏa mãn)
c, Để \(\dfrac{2x+1}{x+3}\Leftrightarrow2x+1⋮x+3\)
<=> 2(x + 1) - 1 \(⋮x+3\)
<=> -1 \(⋮x+3\)
=> x + 3 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> x = -2; -4 (thỏa mãn)
@Nguyễn Bá Minh

Answer 2

2. a, (x - 1)(y + 2) = -7
Do x; y \(\in Z\Rightarrow x-1;y+2\in Z\)
Mà (x - 1)(y + 2) = -7
=> x - 1; y + 2 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Nếu \(\left\{{}\begin{matrix}x-1=1\\y+2=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-9\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-1\\y+2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=7\\y+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-7\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-1\end{matrix}\right.\) (thỏa mãn)

Vậy các cặp (x; y) thỏa mãn là (2; -9); (0; 5); (8; -3); (-6; -1)
@Nguyễn Bá Minh

Answer 3

2. b, x(y - 3) = -12
Do x; y \(\in Z\Rightarrow y-3\in Z\)
Mà x(y - 3) = -12
=> x; y - 3 \(\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng :

x	1	-1	2	-2	3	-3	4	-4	6	-6	12	-12
y - 3	-12	12	-6	6	-4	4	-3	3	-2	2	-1	1
y	-9	15	-3	9	-1	7	0	6	1	5	2	4

@Nguyễn Bá Minh

Answer 4

2.

a) \(\left(x-1\right)\left(y+2\right)=-7\)

TH1 : \(\left[{}\begin{matrix}x-1=1\\y+2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=-9\end{matrix}\right.\)

TH2:\(\left[{}\begin{matrix}x-1=-7\\y+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-1\end{matrix}\right.\)

TH3: \(\left[{}\begin{matrix}x-1=-1\\y+2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=5\end{matrix}\right.\)

TH4: \(\left[{}\begin{matrix}x-1=7\\y+2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

b)Tương tự