Question

Tìm các cặp số nguyên (x;y) t/mãn: 4x-5y-6xy+7=0

Answer 1

4x-5y-6xy+7=0

=>4x-6xy-5y+7=0

=>\(6xy-4x+5y-7=0\)

=>\(2x\left(3y-2\right)+5y-\frac{10}{3}-\frac{11}{3}=0\)

=>\(6x\left(y-\frac23\right)+5\left(y-\frac23\right)=\frac{11}{3}\)

=>\(\left(y-\frac23\right)\left(6x+5\right)=\frac{11}{3}\)

=>\(\left(6x+5\right)\left(3y-2\right)=11\)

=>(6x+5;3y-2)∈{(1;11);(11;1);(-1;-11);(-11;-1)}

=>(6x;3y)∈{(-4;13);(6;3);(-6;-9);(-16;1)}

=>(x;y)∈{(-2/3;13/3);(1;1);(-1;-3);(-8/3;1/3)}

mà x,y nguyên

nên (x;y)∈{(1;1);(-1;-3)}