Tìm `x` biết: `a)` `1/3x + 2/5 (x-1) = 0` `b)` `(2x-3)-(x+2)=(x-2)-3(x-5)`

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1.  `a)1/3*x+2/5(x-1)=0`

   `⇒1/3*x+2/5*x+2/5*(-1)=0`

   `⇒1/3*x+2/5*x-2/5=0`

   `⇒x*(1/3+2/5)-2/5=0`

   `⇒x*11/15-2/5=0`

   `⇒x*11/15=2/5`

   `⇒x=2/5÷11/15`

   `⇒x=6/11`

    

   `b)(2x-3) – (x+2) = (x-2) – 3(x-5)`

   `⇒ 2x -3 – x -2 = x-2 – 3x + 15`

   `⇒ (2x-x) – (3+2) = (x-3x) – (2-15)`

   `⇒ x – 5 = -2x + 13`

   `⇒ x + 2x= 13 + 5`

   `⇒ 3x= 18`

   `⇒ x= 18:3`

   `⇒ x=6`

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2. a) `1/3x + 2/5(x-1) =0`

   `=> 1/3 x + 2/5x – 2/5 =0`

   `=> x(1/3 + 2/5) – 2/5 =0`

   `=> x. 11/15 – 2/5 =0`

   `=>x. 11/15 = 2/5`

   `=> x= 2/5 : 11/15`

   `=> x = 2/5. 15/11`

   `=> x= 6/11`

   Vậy `x= 6/11`

   b) `(2x-3) – (x+2) = (x-2) – 3(x-5)`

   `=> 2x -3 – x -2 = x-2 – 3x + 15`

   `=> (2x-x) – (3+2) = (x-3x) – (2-15)`

   `=> x – 5 = -2x + 13`

   `=> x + 2x= 13 + 5`

   `=> 3x= 18`

   `=> x= 18:3`

   `=> x=6`

   Vậy `x=6`

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