←Vrati se na zadatke7. ZadatakRiješite jednadžbu (xx−3)+4⋅(x+1x−2)+(x+23)=125.\binom{x}{x-3}+4\cdot\binom{x+1}{x-2}+\binom{x+2}3=125.(x−3x)+4⋅(x−2x+1)+(3x+2)=125.Prikaži rješenjeRješenjePrimijenimo li svojstvo simetričnosti binomnih koeficijenata, jednadžbu pišemo u obliku (x3)+4⋅(x+13)+(x+23)=125.\binom x3+4\cdot\binom{x+1}3+\binom{x+2}3=125.(3x)+4⋅(3x+1)+(3x+2)=125. x(x−1)(x−2)6+4(x+1)x(x−1)6+(x+2)(x+1)x6=125.\frac{x(x-1)(x-2)}6+4\frac{(x+1)x(x-1)}6+\frac{(x+2)(x+1)x}6=125.6x(x−1)(x−2)+46(x+1)x(x−1)+6(x+2)(x+1)x=125. Nakon provedenog množenja imamo: x⋅(x2−3x+2+4x2−4+x2+3x+2)=6⋅1256x3=6⋅125x=5\begin{aligned} x\cdot(x^2-3x+2+4x^2-4+x^2+3x+2)&=6\cdot125\\ 6x^3&=6\cdot125\\ x&=5 \end{aligned}x⋅(x2−3x+2+4x2−4+x2+3x+2)6x3x=6⋅125=6⋅125=5Pomakni formulu lijevo ili desno.
