Đáp án:
a) (x+1)(2x-1)=0
th1=> x+1=0
=>x=-1
th2: 2x-1 = 0
=> x=1/2
b) x+3/x+1 + x+2/x=2 ( x# -1 , x#0)
x ( x+3)+(x+1).(x+2)/x(x+1)=2
x^2+3x +x^2+2x+x+2/x(x+1)=2
2x^2+6x+2=2x(x+1)
2x^2+6x+2=2x^2+2x
6c+2=2x
6x-2x=-2
4x=-2
x= -1/2 x#-1, x#0
c) x-3/5+1>2x-5
x-3+5>10x-25
x+2>10x-25
x-10x>-25-2
-9x>-27
x<3
d)Ix-3I +5x=9
th1x-3+5x=9, x>=0 x=2, x>=3.. x e rỗng
th2-(x-3)+5x=9, x-3<0 x= 3/2... x e 3/2
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