Question

[ミ★ℒїøŋεℓ Ɱεşşї★彡 (Team Football P...]

tìm các số nguyên x,y (x+y)*(y-2)=5                  x*(y-4)=12                        xy-5x-5y=0

Answer 1

[Đặng Tú Phương]

\(\left(x+y\right)\left(y-2\right)=5\)

\(\Rightarrow\left(x+y\right);\left(y-2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Xét bảng 

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

Vậy.............

\(x\left(y-4\right)=12\)

\(\Rightarrow x;\left(y-4\right)\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Xét bảng

x	1	-1	2	-2	3	-3	4	-4	6	-6	12	-12
y-4	12	-12	6	-6	4	-4	3	-3	2	-2	1	-1
x	1	-1	2	-2	3	-3	4	-4	6	-6	12	-12
y	16	-8	10	-2	8	0	7	1	6	2	5	3

Vậy............................

\(xy-5x-5y=0\)

\(\Rightarrow x\left(y-5\right)-5y+25=0+25\)

\(\Rightarrow x\left(y-5\right)-5\left(y-5\right)=25\)

\(\Rightarrow\left(x-5\right)\left(y-5\right)=25\)

\(\Rightarrow\left(x-5\right)\left(y-5\right)\inƯ\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

Xét bảng

x-5	1	-1	5	-5	25	-25
y-5	25	-25	5	-5	1	-1
x	6	4	10	0	30	-20
y	30	-20	10	0	6	4

Vậy......................................

Answer 2

[ミ★ℒїøŋεℓ Ɱεşşї★彡 (Team Football P...]

Thanks bn :))