Question

bài 1 Tìm các số nguyên x, y biết:

a) (x + 1).(y - 2) = 5                  

b) (x - 5).(y + 4) = -7          

c) (x + 1)² + (y – 1)² = 0              

d) (2x – 18)² + ( y + 37)² = 0

e) x-(17-8)=5+(10-3x)   

Answer 1

a) 

\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Ta có bảng:

x+1	1	-1	5	-5
y-2	5	-5	1	-1
x	0	-2	4	-6
y	7	-3	3	1

Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)

 

 

Answer 2

b) 

\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

Ta có bảng:

x-5	1	-1	7	-7
y+4	-7	7	-1	1
x	6	4	12	-2
y	-11	3	-5	-3

Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)

 

Answer 3

e) 

\(x-\left(17-8\right)=5+\left(10-3x\right)\\ \Rightarrow x-9=5+10-3x\\ \Rightarrow x+3x=5+10+9\\ \Rightarrow4x=24\\ \Rightarrow x=\dfrac{24}{4}=6\)

Vậy \(x=6\)

Answer 4

Bài 1: 

a) Ta có: (x+1)(y-2)=5

nên x+1;y-2 là các ước của 5

\(\Leftrightarrow x+1;y-2\inƯ\left(5\right)\)

\(\Leftrightarrow x+1;y-2\in\left\{1;-1;5;-5\right\}\)

Trường hợp 1:

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

Trường hợp 2: 

\(\left\{{}\begin{matrix}x+1=5\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}x+1=-1\\y-2=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

Trường hợp 4:

\(\left\{{}\begin{matrix}x+1=-5\\y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)

b) Ta có: \(\left(x-5\right)\left(y+4\right)=-7\)

nên x-5;y+4 là các ước của -7

\(\Leftrightarrow x-5;y+4\inƯ\left(-7\right)\)

\(\Leftrightarrow x-5;y+4\in\left\{1;-1;7;-7\right\}\)

Trường hợp 1:

\(\left\{{}\begin{matrix}x-5=1\\y+4=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-11\end{matrix}\right.\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}x-5=-7\\y+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}x-5=-1\\y+4=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=3\end{matrix}\right.\)

Trường hợp 4: 

\(\left\{{}\begin{matrix}x-5=7\\y+4=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y=-5\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(6;-11\right);\left(-2;-3\right);\left(4;3\right);\left(12;-5\right)\right\}\)

c) Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Vậy: (x,y)=(-1;1)

d) Ta có: \(\left(2x-18\right)^2\ge0\forall x\)

\(\left(y+37\right)^2\ge0\forall y\)

Do đó: \(\left(2x-18\right)^2+\left(y+37\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}2x-18=0\\y+37=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=18\\y=0-37\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=-37\end{matrix}\right.\)

Vậy: (x,y)=(9;-37)

e) Ta có: \(x-\left(17-8\right)=5+\left(10-3x\right)\)

\(\Leftrightarrow x-17+8=5+10-3x\)

\(\Leftrightarrow x-9=-3x+15\)

\(\Leftrightarrow x-9+3x-15=0\)

\(\Leftrightarrow4x-24=0\)

\(\Leftrightarrow4x=24\)

hay x=6

Vậy: x=6