a.x(y+3)=3
=> x(y+3) ∈Ư(3)={-3;-1;1;3}
ta có bảng sau
| x | -3 | -1 | 1 | 3 |
| y+3 | -1 | -3 | 3 | 1 |
| y | -4 | -6 | 0 | -2 |
vậy x=-3 thì y=-4
x=-1 thì y=-6
x=1 thì y=0
x=3 thì y=-2
c.x+3⋮ x+1
=> (x+3)-(x+1)⋮(x+1)
=> (x+3-x-1)⋮(x+1)
=> 2⋮(x+1)
=> (x+1) ∈ Ư(2)={-2;-1;1;2}
=> x∈{-3;-2;0;1}
vậy x ∈{-3;-2;0;1}
b,d tương tự
a.(x-2)(x+3)>0
=>\(\left[{}\begin{matrix}x-2>0\\x+3>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x>-3\end{matrix}\right.\)
=> x>2
vậy x>2
b.(x-2)(x-1)>0
=> \(\left[{}\begin{matrix}x-2>0\\x-1>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x>1\end{matrix}\right.\)
=> x>2
vậy x>2
c.(x-2)(x2+1)>0
=> \(\left[{}\begin{matrix}x-2>0\\x^2+1>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x^2>-1\Rightarrow x>\sqrt{-1}\end{matrix}\right.\)
vậy x>2
d.(x-1)(x+2)>0
=> \(\left[{}\begin{matrix}x-1>0\\x+2>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>1\\x>-2\end{matrix}\right.\)
=> x>1
vậy x>1
Còn câu này nx bn ạ:
x^2.(x+2)<0
Tìm x
Giúp mk nhanh nha, mk cần gấp
