Giải phương trình: $\frac{\sqrt[]{x – 2009} – 1}{x – 2009}$ + $\frac{\sqrt[]{y – 2010} – 1}{y – 2010}$ + $\frac{\sqrt[]{z – 2011} – 1}{z – 2011}$ = $

Đáp án:

`(\sqrt{x – 2009} – 1)/(x – 2009) +  (\sqrt{y – 2010} – 1)/(y – 2020) + (\sqrt{z – 2011} – 1)/(z – 2011) = 3/4`

Đặt : `\sqrt{x – 2019} = a`

        `\sqrt{y – 2010} = b`

         `\sqrt{z – 2011} = c`

phương trinh sẽ thành :

`(a – 1)/a^2 + (b – 1)/b^2 + (c – 1)/c^2 = 3/4`

`<=> 1/a – 1/a^2 + 1/b – 1/b^2 + 1/c – 1/c^2 = 3/4`

`<=> 1/a – 1/a^2 + 1/b – 1/b^2 + 1/c – 1/c^2 – 3/4 = 0`

`<=> (1/a^2 – 2. 1/a . 1/2 + 1/4) + (1/b^2 – 2. 1/b . 1/2 + 1/4) + (1/c^2 – 2 . 1/c . 1/2 + 1/4) = 0`

`<=> (1/a – 1/2)^2 + (1/b – 1/2)^2 + (1/c – 1/2)^2 = 0`

Do ` (1/a – 1/2)^2 ≥ 0` 

     `(1/b – 1/2)^2 ≥ 0`

      `(1/c – 1/2)^2 ≥ 0`

`=> (1/a – 1/2)^2 + (1/b – 1/2)^2 + (1/c – 1/2)^2 ≥ 0`

Dấu “=” xẩy ra

<=> $\left[ \begin{array}{l}\dfrac{1}{a} – \dfrac{1}{2} = 0 \\\dfrac{1}{b} – \dfrac{1}{2} = 0\\\dfrac{1}{c} – \dfrac{1}{2} = 0\end{array} \right.$

<=>$ \left[ \begin{array}{l}\\\sqrt{x – 2009} = 2\\\sqrt{y – 2010} = 2\\\sqrt{z – 2011} = 2\end{array} \right.$

<=>$ \left[ \begin{array}{l}x =2013\\b= 2014\\z = 2015\end{array} \right.$

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