Tìm đa thức bậc ba $f\left( x \right) = a{x^3} + b{x^2} + cx + 1$ (với a ≠ 0 ) biết $f\left( {-1} \right) = \;-2,{\rm{ }}f\left( 1 \right) = 2,{\rm{ }}f\left( 2 \right) = 7.$

A. $I = \dfrac{{40}}{3}\left( A \right),\,{I_1} = \dfrac{{80}}{3}\left( A \right),\,{I_2} = \dfrac{{100}}{3}\left( A \right)$.

B. $I = \dfrac{{80}}{7}\left( A \right),\,{I_1} = \dfrac{{40}}{7}\left( A \right),\,{I_2} = \dfrac{{100}}{7}\left( A \right)$.

C. $I = \dfrac{{40}}{11}\left( A \right),\,{I_1} = \dfrac{{80}}{11}\left( A \right),\,{I_2} = \dfrac{{120}}{11}\left( A \right)$.

D. $I = \dfrac{{40}}{7}\left( A \right),\,{I_1} = \dfrac{{80}}{7}\left( A \right),\,{I_2} = \dfrac{{120}}{7}\left( A \right)$.

A. $I = \dfrac{8}{{15}},{I_2} = \dfrac{4}{{15}},{I_3} = \dfrac{4}{{15}}.$

B. $I = \dfrac{8}{7},{I_2} = \dfrac{4}{7},{I_3} = \dfrac{4}{7}.$

C. $I = \dfrac{8}{5},{I_2} = \dfrac{4}{5},{I_3} = \dfrac{4}{5}.$

D. $I = \dfrac{8}{{11}},{I_2} = \dfrac{4}{{11}},{I_3} = \dfrac{4}{{11}}.$

24Ω và 12Ω

12Ω và 24Ω

2Ω và 1Ω

3Ω và 2Ω

$I = \dfrac{{16}}{{11}}\left( A \right),\,{I_1} = \dfrac{8}{{11}}\left( A \right),\,{I_2} = \dfrac{8}{{11}}\left( A \right).$

$I = \dfrac{8}{{11}}\left( A \right),\,{I_1} = \dfrac{{16}}{{11}}\left( A \right),\,{I_2} = \dfrac{{16}}{{11}}\left( A \right).$

$I = \dfrac{{24}}{7}\left( A \right),\,{I_1} = \dfrac{8}{7}\left( A \right),\,{I_2} = \dfrac{{16}}{7}\left( A \right).$

$I = \dfrac{{24}}{{11}}\left( A \right),\,{I_1} = \dfrac{8}{{11}}\left( A \right),\,{I_2} = \dfrac{{16}}{{11}}\left( A \right).$

A. 32, 30, 34

B. 32, 33, 31

C. 32, 34, 30 

D. 31, 32, 30 

A.250 g, 150 g, 625 g.

B.125 g, 250 g, 625 g.

C.125 g, 350 g, 525 g.

D.250 g, 125 g, 625 g.