a)$\sqrt{3x+1}$-$\sqrt{x-1}$ =2
($\sqrt{3x+1}$-$\sqrt{x-1}$ )²=2
3x+1-2$\sqrt{(3x+1)(x-1)}$+x+1=2
4x=2$\sqrt{(3x+1)(x-1)}$
(4x)²=(2$\sqrt{(3x+1)(x-1)}$)²
16.x²=4.(3x+1)(x-1)
16.x²=(3x+1)(4x-4)
16.x²=12.x²+4x-12x-4
16.x²=12.x²-8x-4
4x²+8x+4=0
x²+2x+1=0
(x+1)²=0
x=-1
b) đặt $\sqrt{x}$ =a
ta có phương trình
$\frac{a²-4}{a+2}$=a²-8
$\frac{(a-2)(a+2)}{a+2}$=a²-8
a-2=a²-8
a²-8-a+2=0
a²-a-6=0
a²-3a+2a-6=0
(a-3)(a+2)=0
⇒a=3(a>0)
$\sqrt{x}$ =3
x=9
c) vì $\sqrt{x}$ +2 >0 nên
$\frac{2}{\sqrt{x}+2}$> $\frac{1}{2}$
4>$\sqrt{x}$ +2
2>$\sqrt{x}$
0≤x<4
d,e minh chụi
