(7-x)(2x-4)(x2+2)=0
mà \(x^2+2>=2>0\forall x\)
nên (7-x)(2x-4)=0
=>\(\left[{}\begin{matrix}7-x=0\\2x-4=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
1,-15(x+2)+7(2x-3)
2,x^2+2x=0
3,(-2x)(-4x)+28=100
4,(2x-5)^2-49=0
5,/x-5/+/y-3/=1
6,3×(x-2)-6(x-5)=2(5-2x)
7, 5x ×(-x)^2+1=0
8,4x^3=4x
1, (x - 4/7) : (x+1/2) > 0
2, (2x - 3) : (x+7/4) < 0
Tìm x
1)7.(x-1) +2x.(x-1) =0
2)2x.(x-3)+9x-27=0
3)4.(x-1/2)+4.(5-x)=9/2
Tìm x, y thuộc Z
a) 3 (x + 2) + 2 (x - 3) = 5x - 7
b) 4(2x+1)-7(x-2)-x=15
c) (x-3)(2x+10)=0
d) (7-x)(x+3)>0
e) (x2+7)(x2-7)<0
f) (2x-3)(2y+3)=17
g) (2x-3)(y+3)=24
( x - 4/5 ).( x+2 và 1/5 )= 0
x+20%x=12
1/4 - (2x+ 1/2 )^2=0
(x-1/5)^2 + 1= 3,5 : 7%
tìm x
7.(x+2)-4.(x-1)=30
(2x -2).(x-4)=0
(2x -1).(x-2)=0
help me
tìm x thuộc Z biết :
a)|3x+5|-2(x+7)-4(7-x)=|3x+5|+2x
b)|x^2-4|+(x-2)^2=0
c)|x+7|+|x+5|+|x+3|=2x
d)|3x+5|=5-3x
Tìm x
(2x-7)+17=6
12-2.(3-3x)=-2
-14+3.(-x+5)=-20
-90:5.(-3-2x)=6
(x+1).(x-3)=0
(2x-2).(x+4)=0
(22+4).(x+3)=0
(5-x).(6-2x)=0
3.(x+1)+5=x+8
-4.(2x+9)-(-8x+3)-(x+13)=0
