x-1/2+x-1/3+x-1/6=2 help

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1. `(x-1)/2+(x-1)/3+(x-1)/6=2`

   `⇒[3.(x-1)+2.(x-1)+(x-1)]/6=2`

   `⇒[3.(x-1)+2.(x-1)+(x-1)]/6=12/6`

   `⇒3.(x-1)+2.(x-1)+(x-1)=12`

   `⇒3.x-3.1+2.x-2.1+x-1=12`

   `⇒3x-3+2x-2+x-1=12`

   `⇒(3x+2x+x)-(3+2+1)=12`

   `⇒6x-6=12`

   `⇒6(x-1)=12`

   `⇒x-1=12÷6`

   `⇒x-1=2`

   `⇒x=2+1`

   `⇒x=3`

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2. `(x-1)/2+(x-1)/3+(x-1)/6=2`

   `⇔(3(x-1)+2(x-1)+(x-1))/6=12/6`

   `⇔3x-3+2x-2-x+1=12`

   `⇔6x=12`

   `⇔x=3`

   Vậy `x=3`

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