Cho hai đa thức: U(x) = 3(x^4)-5(x^3) + x^2 - 3/2x - 6; V(x) = - 6(x^2) - 5x + 3. a) Tính U( - 1) và V(1/2); b) Tìm đa thức Z(x) biết Z(x) = 2U(x) + V(x).
a) Ta có \[U( - 1) = \;3\,\,.\,\,{( - 1)^4}--5\,\,.\,\,{( - 1)^3} + {( - 1)^2} - \frac{3}{2}\,.\,\,( - 1) - 6\]
\[ = \;3\,\,.\,\,1--5\,\,.\,\,( - 1) + 1 + \frac{3}{2} - 6 = \;3 + 5 + 1 + \frac{3}{2} - 6 = 3 + \frac{3}{2} = \frac{9}{2}\];
\[V\left( {\frac{1}{2}} \right) = - 6\,\,.\,\,{\left( {\frac{1}{2}} \right)^2} - 5\,\,.\,\,\left( {\frac{1}{2}} \right) + 3 = - 6\,\,.\,\,\frac{1}{4} - \frac{5}{2} + 3 = \frac{{ - 3}}{4} - \frac{5}{2} + 3 = \frac{{ - 1}}{4}\].
Vậy \(U( - 1) = \frac{9}{2}\) và \[V\left( {\frac{1}{2}} \right) = \frac{{ - 1}}{4}\].
b) Ta có \[Z(x) = 2U(x) + V(x) = 2\,\,.\,\,\left( {\;3{x^4}--5{x^3} + {x^2} - \frac{3}{2}x - 6} \right) + \left( { - 6{x^2} - 5x + 3} \right)\]
\[ = 6{x^4}--10{x^3} + 2{x^2} - 3x - 12 - 6{x^2} - 5x + 3\]
\[ = 6{x^4}--10{x^3} + \left( {2{x^2} - 6{x^2}} \right) - \left( {3x + 5x} \right) + \left( {3 - 12} \right)\]
\[ = 6{x^4}--10{x^3} - 4{x^2} - 8x - 9\].
Vậy \[Z(x) = 6{x^4}--10{x^3} - 4{x^2} - 8x - 9\].