ta có 2xy + x + y = 21
=> 4xy + 2x + 2y = 42
=> (4xy + 2x) + 2y = 42
=> 2x(2y+1) + 2y + 1 = 43
=> (2x + 1)(2y+1) = 43
\(\Rightarrow\hept{\begin{cases}2x+1\in\left\{1;43;-43;-1\right\}\\2y+1\in\left\{43;1;-1;-43\right\}\end{cases}}\Rightarrow\hept{\begin{cases}x\in\left\{0;21;-22;-1\right\}\\y\in\left\{21;0;-1;-22\right\}\end{cases}}\)
tim x, y thuoc Z, biet:
x - y + 2xy bang 7
a,tim x biet |x-2|+|3-2x|=2x+1
b,tim x,y thuoc Z biet xy+2x-y=5
c, tinh A=(1-1/15)(1-1/21)(1-1/28).....(1-1/210)
tim x y thuoc z biet (x+y).2=xy
tim x,y thuoc Z biết 8y-x=2xy
tim x,y thuoc z biet x/5=y/3=z/4 và x-z=7
tim x,y thuộc z biết x/3=y/4=z/5 và 2z+3y+5z=86
tim cac so x,y,z thuoc Q biet rang (x+y):(5-z):(y+z):(y+9)=3:1:2:5
tim x,y thuoc Z biet 25-y^2=9(x-2018)^2
tim x;y thuoc z biet:
1/x + 1/y = 1/5
tim x thuoc z biet y^2-x^2=102
