Có: y = $\sqrt[]{2x-x²}$
=> y' = $\frac{(2x-x²)'}{2\sqrt[]{2x-x²}}$
= $\frac{2-2x}{2\sqrt[]{2x-x²}}$
= $\frac{2(1-x)}{2\sqrt[]{2x-x²}}$
= $\frac{1-x}{\sqrt[]{2x-x²}}$
=> y'' = $\frac{(1-x)'(\sqrt[]{2x-x^{2}})-(1-x)(\sqrt[]{2x-x^{2}})'}{2x-x^{2}}$
= $\frac{-\sqrt[]{2x-x^{2}}-\frac{(1-x)^{2}}{\sqrt[]{2x-x^{2}}}}{2x-x^{2}}$
= $\frac{\frac{-(2x-x^{2})-(1-x^{2})}{\sqrt[]{2x-x^{2}}}}{2x-x^{2}}$
= $\frac{\frac{-2x+x^{2}-1+2x-x^{2}}{\sqrt[]{2x-x^{2}}}}{2x-x^{2}}$
= $\frac{-1}{(2x-x^{2})\sqrt[]{2x-x^{2}}}$
=> A = ($\sqrt[]{2x-x^{2}}$)³.$\frac{-1}{(2x-x^{2})\sqrt[]{2x-x^{2}}}$ + 1
= $\frac{-(\sqrt[]{2x-x^{2}})³}{(2x-x^{2})\sqrt[]{2x-x^{2}}}$ + 1
= $\frac{-(\sqrt[]{2x-x^{2}})²}{(2x-x^{2})}$ + 1
= -1 + 1
= 0
Vậy A = 0.
