Tìm x: a)5 √x-2=x+2 b) √x ²-2x+1=5

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1. Đáp án:

   5$\sqrt{x-2}$ = `x` `+` `2`

   `⇒` $(5\sqrt{x-2}$ )$^{2}$ `=` `(x+2)^2`

   `⇒` `25` `×`  `(x-2)^2` = `x^2` `+` `4x` `+` `4`

   `⇒` `25` `=` `(x+2)^2` `÷` `(x-2)^2`

   `⇒` `5^2` `=` ($\frac{x+2}{x-2}$)$^{2}$

   `⇒` $\frac{x+2}{x-2}$ `=` `5`

   `⇒` `x` `+` `2` `=` `5x` `-` `10`

   `⇒` `4x` `=` `12`

   `⇒` `x` `=` `3`

   b,

   $\sqrt(x^2 – 2x+1)$ = `5` (x ≥ 1)

   `<=>` $\sqrt((x-1)^2)$ `=` `5`

   `=>` `x` – `1` `=` `5`

   `<=>` `x` `=` `6`

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2. a) `5\sqrt(x-2) = x+2 (x≥2)`

   `=> 25(x-2) = (x+2)^2`

   `<=> 25x – 50 = x^2 + 4x + 4`

   `<=> x^2 -21x + 54 = 0`

   `<=> x=18` hoặc `x=3`

   Vậy `S = {3;18}`

   b) `\sqrt(x^2 – 2x+1) = 5 (x ≥ 1)`

   `<=> \sqrt((x-1)^2) = 5`

   `=> x-1=5`

   `<=> x = 6`

   Vậy `S = {6}`

    

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