Question

gasiti termenul care nu contine x in dezvoltarea binomului (x la 2+1/x)la puterea 12

Answer 1

Formula  termenului  liber  este 
T(k+1)=C12^kx^(12-k)*(1/x)^K  unde C12^k= combinari  de  12  luate  cate  k
Determini  pe  k
x^(12-k)*(1/x)^k  =x^(12-2k).  Pui  conditia  ca  12-2k=0  Pt  ca  x^0=1
12-2k=0  k=6
T(6+1)=C12^6 
T7=12!/6!*(12-6)!=7*8*9*10*11*12/6!
6!=1*2*3*4*5*6 =720
Calculwele  ti  le  faci  singur

Answer 2

$T_{k+1}= C^{k} _{12}*( x^2)^{12-k} *( \frac{1}{x})^k \\ x^{24-2k-k} =x^0 \\ 24-3k=0 \\ 24=3k \\ k=8 \\ T_{8+1}= C^{8} _{12}* x^{24-16}* x^{-8} = \frac{12!}{8!*4!}= \frac{9*10*11*12}{24}=495$