a,
đkxđ : x khác + - 2 ; 1/3
=> \(3x-1=x-2\)
<=> 3x - 1 - x + 2 = 0
=> 2x +1 =0
=> x = -1/2 ( tm)
`a)\sqrt{(3x-1)^2}=\sqrt{(x-2)^2}`
`<=>(3x-1)^2=(x-2)^2`
`<=>(3x-1)^2-(x-2)^2=0`
`<=>(3x-1-x+2)(3x-1+x-2)=0`
`<=>(2x+1)(4x-3)=0`
`<=>`$\left[\begin{matrix} x=\dfrac{-1}{2}\\ x=\dfrac{3}{4}\end{matrix}\right.$
Vậy `S={[-1]/2;3/4}`
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`b)\sqrt{x^2+4x+4}=\sqrt{x^2-6x+9}`
`<=>\sqrt{(x+2)^2}=\sqrt{(x-3)^2}`
`<=>(x+2)^2=(x-3)^2`
`<=>(x+2)^2-(x-3)^2=0`
`<=>(x+2-x+3)(x+2+x-3)=0`
`<=>6(2x-1)=0`
`<=>2x-1=0<=>x=1/2`
Vậy `S={1/2}`
b ) .
<=> \(x^2+4x+4=x^2-6x+9\)
=> \(x^2+4x+4-x^2+6x-9=0\)
=> \(10x-5=0\)
=> x = 5/10 = 1/2
