(1,0 điểm) Cho F ( x ) = 5 x 2 − 1 + 3 x + x 2 − 5 x 3 và G ( x ) = 2 − 3 x 3 + 6 x 2 + 5 x − 2 x 3 − x . a) Thu gọn đa thức F ( x ) , G ( x ) và tính F ( x ) − G ( x ) . b) T

Hướng dẫn giải

a) Ta có: \(F\left( x \right) = 5{x^2} - 1 + 3x + {x^2} - 5{x^3}\)

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

\(F\left( x \right) = - 5{x^3} + 6{x^2} + 3x - 1\).

Ta có: \(G\left( x \right) = 2 - 3{x^3} + 6{x^2} + 5x - 2{x^3} - x\)

\(G\left( x \right) = \left( { - 3{x^3} - 2{x^3}} \right) + 6{x^2} + \left( {5x - x} \right) + 2\)

\(G\left( x \right) = - 5{x^3} + 6{x^2} + 4x + 2\).

Ta có: \(F\left( x \right) - G\left( x \right) = - 5{x^3} + 6{x^2} + 3x - 1 - \left( { - 5{x^3} + 6{x^2} + 4x + 2} \right)\)

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

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

\(F\left( x \right) - G\left( x \right) = - x - 3\).

b) Ta có: \(N\left( x \right) + F\left( x \right) = - G\left( x \right)\)

Do đó, \(N\left( x \right) = - G\left( x \right) - F\left( x \right)\)

Suy ra \(N\left( x \right) = - \left( { - 5{x^3} + 6{x^2} + 4x + 2} \right) - \left( { - 5{x^3} + 6{x^2} + 3x - 1} \right)\)

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

\(N\left( x \right) = 10{x^3} - 12{x^2} - 7x - 1\).