Ta có : `a + b = 7`
`a + b = 7 => ( a+ b )^2 = 49`
`<=> a^2 + 2ab + b^2 = 49`
`<=> a^2 + b^2 = 25`
`<=> a^2 - 2ab + b^2 = 25 - 2ab`
`<=> ( a - b )^2 = 25 - 2.12`
`<=> ( a - b )^2 = 1`
`<=>` $\left[ \begin{array}{l}a-b=1\\a-b=-1\end{array} \right.$
`<=>` $\left[ \begin{array}{l}(a-b)^{2009} =1\\(a-b)^{2009} =-1\end{array} \right.$
Vậy `(a-b)^{2009}` `= -1` hoặc `(a-b)^{2009}` `- 1`
