VÌ \(\left|x\right|\ge0;\left|y\right|\ge0;\left|z\right|\ge0\)NÊN ĐỂ\(\left|x\right|+\left|y\right|+\left|z\right|=0\)\(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\\\left|z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}}\)
tim x,y,z thuoc Z biet /2x-4/+/y+2/+/2x+3y-z/=0
tim x, y thuoc z biet :
/ x-y / + / y-z/ + /z - x / = 2015
Tim x, y thuoc Z biet : | x-2| + (x-y+1)^2 =0
tim x,y thuoc z biet (x-2).(y+12)<0
cho P =|x|.(y-1).Tim x,y thuoc Z biet P<0
Cho A thuoc {0.1.2.3.4.5.6.7.8.9.10} biet A co hon 1 phan tu neu x,y,z thuoc A va x<y<z thi z =x.y tim A
tim x,y thuoc Z biet x*y=x+y
Tim cap x,y biet x,y thuoc Z va
2xy- 4+2x+y=0
tim x,y thuoc z biet
-24/-6 = x/3 = 4/y^2 = z^3/-2
