a) $\dfrac{\left(x+1\right)^2\div 2+1}{2}=600$
$\dfrac{\frac{\left(x+1\right)^2}{2+1}}{2}=600$
$\dfrac{2\cdot \dfrac{\left(x+1\right)^2}{2+1}}{2}=600\cdot \:2$
$\frac{\left(x+1\right)^2}{3}=1200$
$\left(x+1\right)^2=3600$
`=>x∈{56;-61}`
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$\dfrac{\left(-x+2\right)\left[\left(-x-2\right)\div 2+1\right]}{2}=-2000$
$\dfrac{\left(-x+2\right)\left(\frac{\left(-x-2\right)}{2+1}\right)}{2}=-2000$
$\dfrac{\left(-x-2\right)\left(-x+2\right)}{3}=-4000$
$\dfrac{x^2}{3}-\frac{4}{3}+4000=-4000+400$
Vậy không tồn tại `x` thỏa mãn.
