x=6

y = 10

Vì x+y=16 ta có

3+6/5+10=9/15=3/5   6+10=16

Vậy x=6

       y=10

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a: \(\Leftrightarrow\left(x+y;y-5\right)\in\left\{\left(1;16\right);\left(2;8\right);\left(4;4\right);\left(8;2\right);\left(16;1\right);\left(-4;-4\right);\left(-8;-2\right);\left(-16;-1\right)\right\}\)

\(\Leftrightarrow\left(x+y;y\right)\in\left\{\left(1;21\right);\left(2;13\right);\left(4;9\right);\left(8;7\right);\left(16;6\right);\left(-4;1\right);\left(-8;3\right);\left(-16;4\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;7\right);\left(10;6\right)\right\}\)

b: \(\Leftrightarrow\left(2x+3\right)\left(2y-1\right)=10\)

\(\Leftrightarrow\left(2x+3;2y-1\right)\in\left\{\left(1;10\right);\left(10;1\right);\left(2;5\right);\left(5;2\right);\left(-1;-10\right);\left(-10;-1\right);\left(-2;-5\right);\left(-5;-2\right)\right\}\)

mà 2x+3;2y-1 đều là các số lẻ

nên \(\left(x,y\right)\in\varnothing\)

a: \(\Leftrightarrow\left(x+y;y-5\right)\in\left\{\left(1;16\right);\left(2;8\right);\left(4;4\right);\left(8;2\right);\left(16;1\right);\left(-4;-4\right);\left(-8;-2\right);\left(-16;-1\right)\right\}\)

\(\Leftrightarrow\left(x+y;y\right)\in\left\{\left(1;21\right);\left(2;13\right);\left(4;9\right);\left(8;7\right);\left(16;6\right);\left(-4;1\right);\left(-8;3\right);\left(-16;4\right)\right\}\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;7\right);\left(10;6\right)\right\}\)

b: \(\Leftrightarrow\left(2x+3\right)\left(2y-1\right)=10\)

\(\Leftrightarrow\left(2x+3;2y-1\right)\in\left\{\left(1;10\right);\left(10;1\right);\left(2;5\right);\left(5;2\right);\left(-1;-10\right);\left(-10;-1\right);\left(-2;-5\right);\left(-5;-2\right)\right\}\)

mà 2x+3;2y-1 đều là các số lẻ

nên \(\left(x,y\right)\in\varnothing\)

a.

Ta có : ( x + y ) . ( y – 5 ) = 16 = 1 . 16 = 16 . 1 = 2 . 8 = 8 . 2 = 4 . 4

Ta có bảng sau :

Vậy : ( x ; y ) là ( 10 ;6) , ( 1 ; 7 )

a/

\(x+6y⋮17\Rightarrow5\left(x+6y\right)=5x+30y⋮17\)

\(5x+47y=\left(5x+30y\right)+17y\)

\(5x+30y⋮17\left(cmt\right);17y⋮17\Rightarrow5x+47y⋮17\)

b/

\(3x+16y⋮5\Rightarrow2\left(3x+16y\right)=6x+32y=\left(5x+30y\right)+\left(x+2y\right)⋮5\)

Mà \(5x+30y⋮5\Rightarrow x+2y⋮5\)