*Ta có: S = S 1 ⇔ π .x. 3 (1+x) = π . 3 (x+1)
⇔ x(1+x) = x +1 ⇔ x 2 -1 =0 ⇔ (x+1)(x-1) = 0
Vì x > 0 nên x+1 > 0
suy ra: x-1 = 0 ⇔ x = 1
*Ta có: S = 2 S 1 ⇔ π.x. 3 (1+x) = 2.π. 3 (x+1)
⇔ x(x+1) = 2(x+1) ⇔ x 2 – x -2 =0
⇔ x 2 – 2x +x - 2 = 0 ⇔ (x+1)(x-2) = 0
Vì x > 0 nên x+1 > 0
suy ra : x-2 = 0 ⇔ x = 2