Cho f(x)=x^5−3x^2+2x−1 và g(x)=−x^5+4x−5x^3+2=−x^5−5x^3+4x+2.Tìm đa thức h(x) sao cho: (a) f(x)+h(x)=g(x). (b) g(x)+h(x)=f(x).

Ta có \[g\left( x \right) = - {x^5} + 4x - 5{x^3} + 2 = - {x^5} - 5{x^3} + 4x + 2\].

a) Theo đề bài, \[f\left( x \right) + h\left( x \right) = g\left( x \right)\].

Suy ra \[h\left( x \right) = g\left( x \right) - f\left( x \right)\]

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

\[ = - {x^5} - 5{x^3} + 4x + 2 - {x^5} + 3{x^3} - 2x + 1\]

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

\[ = - 2{x^5} - 2{x^3} + 2x + 3\].

Vậy \[h\left( x \right) = - 2{x^5} - 2{x^3} + 2x + 3\].

b) Theo đề bài, \[g\left( x \right) + h\left( x \right) = f\left( x \right)\]

Suy ra \[h\left( x \right) = f\left( x \right) - g\left( x \right)\]

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

\[ = {x^5} - 3{x^3} + 2x - 1 + {x^5} + 5{x^3} - 4x - 2\]

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

\[ = 2{x^5} + 2{x^3} - 2x - 3\].

Vậy \[h\left( x \right) = 2{x^5} + 2{x^3} - 2x - 3.\]