l(x+1)³-(x-2)³l=9
lx³+3x²+3x²-1³-x³-3x²+3x²-1³l=9
l6x²-2l=9
-9≤6x²-2≤9 l + 2
-11≤6x²≤11 l :6
[tex] \frac{-11}{6} [/tex] ≤6≤[tex] \frac{+11}{6} [/tex]
x∈[tex][ \sqrt{ \frac{-11}{6} } , \sqrt \frac{11}{6} ][/tex]
Intr-un fel e bn.. intr-alt fel.. sper sa fie bine..
