Tìm đa thức h( x ) biết f( x ) - h( x ) = g( x ) biết \(f( x ) = 5x - 23x^+ 2x^2 + 1; g( x ) = \frac{1}{3} - \frac{2}{3}x^3 + 2x^2 + x. \)

Ta có: \( f\left( x \right) - h\left( x \right) = g\left( x \right) \Rightarrow h\left( x \right) = f\left( x \right) - g\left( x \right)\)

Mà 

\(\begin{array}{l} f\left( x \right) = 5x - 2{x^3} + 2{x^2} + 1;{\mkern 1mu} {\mkern 1mu} g\left( x \right) = \frac{1}{3} - \frac{2}{3}{x^3} + 2{x^2} + x\\ \to h\left( x \right) = (5x - 2{x^3} + 2{x^2} + 1) - \left( {\frac{1}{3} - \frac{2}{3}{x^3} + 2{x^2} + x} \right)\\ \begin{array}{*{20}{l}} { = 5x - 2{x^3} + 2{x^2} + 1 - \frac{1}{3} + \frac{2}{3}{x^3} - 2{x^2} - x}\\ { = (5x - x) + \left( { - 2{x^3} + \frac{2}{3}{x^3}} \right) + (2{x^2} - 2{x^2}) + \left( {1 - \frac{1}{3}} \right)}\\ { = 4x - \frac{4}{3}{x^3} + \frac{2}{3}} \end{array} \end{array}\)

Chọn đáp án A