(x-5)(y+1)=6
Ta có bảng
| x-5 | 1 | 2 | 3 | 6 |
| y+1 | 6 | 3 | 2 | 1 |
| x | 6 | 7 | 8 | 11 |
| y | 5 | 2 | 1 | 0 |
Vậy các cặp (x;y) là (6;5) ; (7;2) ; (8;1) ; (11;0)
\(\left(x-5\right)\left(x^2.9\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-5=0\\x^2.9=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x^2=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=5\\x=0\end{cases}}\)
tim x, y biet /x+2y+6/+/3x-y+5/=0
a, tim x€Z biet (x-6) chia het cho (x-5)
b, tim x€Z, y€Z biet (x-1).(xy-5)=5
tim x,y thuoc Z biet
|y|.|2x+3|=8
|2x+4|+|y-3|=0
|x-1|+|2y+7|=3
|x+5|+|2y+6| nho hon hoac bang 0
tim cac so nguyen x,y biet
a)(x+5)2.(y-2)=8
b)(2x+4)10+(xy-y-6)=0
c)|x+y-5|+|x-y-3|=0
d)|(x-1)2-1|2015+4-(y-2)2017
tim x ,y biet 5/x-y/3=1/6
tim(x;y) biet -|2x+4| -|y+5| lon hon bang 0 .Tim cap (x;y)
bai 1 tim cac so nguyen x,y biet
a,x/3=7/y b,x/y=-3/11 c,x/y-1=5/-19
bai 2 tim cac so nguyen x,y,z,t biet
12/-6=x/5=-y/3=z/-17=-t/-9
Tim so nguyen x,y biet:
a. (x+2).(x+5)>0
b.(x-1).(y-2)=12
Tim x biet
(x+1/5)^2+|x-y|^2017=0
