Ta có \[B = {x^9} - {x^7} - {x^6} - {x^5} + {x^4} + {x^3} + {x^2} - 1\]

\( = \left( {{x^9} - {x^7}} \right) - \left( {{x^6} + {x^5}} \right) + \left( {{x^4} + {x^3}} \right) + \left( {{x^2} - 1} \right)\)

\( = {x^7}\left( {{x^2} - 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)\)

\( = {x^7}\left( {x - 1} \right)\left( {x + 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)\)

\( = \left( {x + 1} \right)\left[ {{x^7}\left( {x - 1} \right) - {x^5} + {x^3} + \left( {x - 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - \left( {{x^5} - {x^3}} \right)} \right]\)

\( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {{x^2} - 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {x - 1} \right)\left( {x + 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} + 1} \right) - {x^3}\left( {x + 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^7} + 1 - {x^4} - {x^3}} \right)\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} - {x^4}} \right) - \left( {{x^3} - 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left[ {{x^4}\left( {{x^3} - 1} \right) - \left( {{x^3} - 1} \right)} \right]\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^3} - 1} \right)\left( {{x^4} - 1} \right)\)

\( = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)\)

\( = \left( {x + 1} \right){\left( {x - 1} \right)^2}\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)\)

\( = {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)\).

Do đó \(B = {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)\).

Thay \(x = 1\) vào \(B\) ta được \(B = {\left( {1 + 1} \right)^2}{\left( {1 - 1} \right)^3}\left( {{1^2} + 1} \right)\left( {{1^2} + 1 + 1} \right) = 0\).

Vậy \(B = 0\) tại \(x = 1\).

Thực hiện phép tính: 2/3x^2 y (3xy^2}- 5/2y)