a, (x² + 1)² - (2x + 100)² = 0
⇔ (x² + 1 -2x - 100).(x² + 1 + 2x + 100) = 0
⇔ ( x²-2x-99).(x²+2x+101) = 0 mà x² + 2x +101 > 0 ( x² + 2x + 101 = (x+1)² + 99 >0)
⇔ x² - 2x - 99 = 0 ⇔ x² - 11x + 9x - 99 = 0
⇔ x.(x-11) + 9.(x-11) = 0
⇔ (x+9).(x-11) = 0
⇔ x = -9 hoặc x = 11
c, 3 - $\frac{x}{2007}$ -1 = 2 - $\frac{x}{2008}$ - $\frac{x}{200}$
⇔ 2 - $\frac{x}{2007}$ = 2 - $\frac{x}{2008}$ - $\frac{x}{200}$
⇔ $\frac{x}{2007}$ = $\frac{x}{2008}$ - $\frac{x}{200}$
⇔ x. ($\frac{1}{2007}$ - $\frac{1}{2008}$ + $\frac{1}{200}$ = 0
mà $\frac{1}{2007}$ - $\frac{1}{2008}$ + $\frac{1}{200}$ khác 0
⇒ x = 0
