0,05( $\frac{2x-2}{2009}$ + $\frac{2x}{2010}$ + $\frac{2x + 2}{2011}$ = 3,3 – ( $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x+1}{2011}$ ) Tìm

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1. Đáp án:

    Đặt `(x – 1)/2009 + x/2010 + (x + 1)/2011 = t`

   `pt <=> 0,1t = 3,3 – t`

   `<=> 0,1t + t = 3,3`

   `<=> 1,1t = 3,3`

   `<=> t = 3`

   `<=> (x – 1)/2009 + x/2010 + (x + 1)/2011 = 3`

   `<=> [(x – 1)/2009 – 1] + [ x/2010 – 1] + [ (x + 1)/2011 – 1] = 0`

   `<=> (x – 2010)/2009 + (x – 2010)/2010 + (x – 2010)/2011 = 0`

   `<=> (x – 2010)(1/2009 + 1/2010 + 1/2011) = 0`

   Do `1/2009 + 1/2010 + 1/2011 > 0`

   `<=> x – 2010 = 0`

   `<=> x = 2010`

   Vậy `pt` có nghiệm duy nhất là `x = 2010`

   Giải thích các bước giải:

    

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2. Giải thích các bước giải:

   `0,05` ( $\frac{2x – 2}{2009}$ + $\frac{2x}{2010}$ + $\frac{2x}{2011}$ ) = `3,3` – ($\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ )

   `<=>“0,1`( $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ ) + ( $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ ) = `3,3`

   `<=>“1,1`( $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ ) = `3,3`

   `<=>` $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ = `3`

   `<=>` $\frac{x – 1}{2009}$ + $\frac{x}{2010}$ + $\frac{x + 1}{2011}$ – `3` = `0`

   `<=>` ( $\frac{x – 1}{2009}$ – `1`) + ( $\frac{x}{2010}$ + `1` ) + ( $\frac{x + 1}{2011}$ + `1` ) = `0`

   `<=>` $\frac{x – 2010}{2009}$ + $\frac{x – 2010}{2010}$ + $\frac{x – 2010}{2011}$ = `0`

   `<=>` `( x- 2010 )`( $\frac{1}{2009}$ + $\frac{1}{2010}$ + $\frac{1}{2011}$ ) = `0`

   `<=>` ` x- 2010` = `0`

   `<=>` `x = 2010`

   ` Vậy ` `S` = `{ 2010 }`

    

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