Ta có: `x - 2 < x + 1`
Do `(x+1)(x-2) < 0`
`=>` \(\left[{}\begin{matrix}x-2>0\\x+1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)
Vậy...
Bài 1 : tìm x biết
a) ( /x/ - 1/4 ) . ( x2 - 9 ) = 0
b) ( /x/ + 2 ) . ( /x/ - 4 ) = 0
c) ( x2 - 1/4 ) . ( x2 - 1/10 ) = 0
d) ( x + 2 ) . ( x - 3 ) < 0
e) ( x - 1/4 ) . ( x + 1/2 ) > 0
Bài 1 : Tìm x biết
a) ( /x/ - 1/4 ) . ( x2 - 9 ) = 0
b) ( /x/ + 2 ) . ( /x/ - 4 ) = 0
c) ( x2 - 1/4 ) . ( x2 - 1/16 ) = 0
d) ( x + 2 ) . ( x - 3 ) < 0
e) ( x - 1/4 ) . ( x + 1/2 ) > 0
|x+3| + |x-y| =0
|x-1| + (y-2)2 =0
|x-3| + |x-2y+1| =0
|x-2| + |3x .y-5| =0
|x-2| - 3|2-x| =-10
|x-3|+ |x-y+1|=0
|x-1| + (y+5)2020 =0
tìm x
(x-1).(x+3)<0
(2-1).(x+5) >0
(x-5).(x+1/2)>0
(x+1).(x-3/2)<0
Tìm x biết : x mũ 2 + ( x - 1 ) mũ 2 = 0 , ( x -1 ) . ( x - 5 ) > 0 , ( x + 1 ) . ( x - 2 ) < 0
Tìm x,y biết :
|x-4|+|6-x|=0
|x+1|-|x-2|=0
|x+2|-|x-3|=0
|x-2|-|x-3|=0
|2x-1|-|x+1|=0
4x-6+2x=12
Tìm x, x thuộc Q:
a) (x +1) * (x - 2) < 0
b) (x - 2) * (x + 2/3) > 0
c) x + 1/2 * (x - 1/3) < 0
Bt1 tìm x
a) (3/x+1/5).(x-1/2)=0
b)(x-3/2).(2x+1)>0
c)(x-2).(x-4)<0
d)(3x+1/5).(2x-1)<0
e)(x-1).(3x-5)<0
a) (x+1).(x-3/2)<0
b) (x-2).(x-1/2)>0
a) 5/2 - x + 4/5 = 2/3 + 4/7
b) ( x - 1 ) x ( x + 2 )< 0
c) ( x + 3/5 ) x ( x+ 1 )<0
d) ( x - 1/3 ) x ( x + 2/5 )>0
