Question

4. tìm số tự nhiên x và y, biết rằng:

a) x.(y - 2 ) = 8

b) ( x - 1 ).( y - 2 )= 9

c) ( x + 1 ) . ( y - 2 ) = 15

Mơn nhé ^ ^

Answer 1

\(a,x\left(y-2\right)=8\\ \Rightarrow x;\left(y-2\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(x\)	\(-8\)	\(-4\)	\(-2\)	\(-1\)	\(1\)	\(2\)	\(4\)	\(8\)
\(y-2\)	\(-1\)	\(-2\)	\(-4\)	\(-8\)	\(8\)	\(4\)	\(2\)	\(1\)
\(y\)	\(1\)	\(0\)	\(-2\)	\(-6\)	\(10\)	\(6\)	\(4\)	\(3\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-4;0\right),\left(-2;-2\right),\left(-1;-6\right),\left(2;6\right),\left(4;4\right),\left(8;3\right)\)

 

 

 

 

 

 

 

Answer 2

\(b,\left(x-1\right)\left(y-2\right)=9\\ \Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\)

\(x-1\)	\(-9\)	\(-3\)	\(-1\)	\(1\)	\(3\)	\(9\)
\(y-2\)	\(-1\)	\(-3\)	\(-9\)	\(9\)	\(3\)	\(1\)
\(x\)	\(-8\)	\(-2\)	\(0\)	\(2\)	\(4\)	\(10\)
\(y\)	\(1\)	\(-1\)	\(-7\)	\(11\)	\(5\)	\(3\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-2;-1\right),\left(0;-7\right),\left(2;11\right),\left(4;5\right),\left(10;3\right)\)

 

Answer 3

\(c,\left(x+1\right)\left(y-2\right)=15\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(15\right)=\left\{-15;-1;1;15\right\}\)

\(x+1\)	\(-15\)	\(-1\)	\(1\)	\(15\)
\(y-2\)	\(-1\)	\(-15\)	\(15\)	\(1\)
\(x\)	\(-16\)	\(-2\)	\(0\)	\(14\)
\(y\)	\(1\)	\(-13\)	\(17\)	\(3\)

 

Vậy \(\left(x;y\right)=\left(-16;1\right),\left(-2;-13\right),\left(0;17\right),\left(14;3\right)\)