Bài 7.15 trang 28

Đề bài

Cho hai đa thức \(A\left( x \right) = {x^4} - 5{x^3} + {x^2} + 5x - \dfrac{1}{3};B\left( x \right) = {x^4} - 2{x^3} + {x^2} - 5x - \dfrac{2}{3}.\)

Hãy tính \(A\left( x \right) + B\left( x \right);A\left( x \right) - B\left( x \right)\)

Lời giải chi tiết

a)

\(\begin{array}{l}A\left( x \right) + B\left( x \right)\\ = \left( {{x^4} - 5{x^3} + {x^2} + 5x - \dfrac{1}{3}} \right) + \left( {{x^4} - 2{x^3} + {x^2} - 5x - \dfrac{2}{3}} \right)\\ = {x^4} - 5{x^3} + {x^2} + 5x - \dfrac{1}{3} + {x^4} - 2{x^3} + {x^2} - 5x - \dfrac{2}{3}\\ = \left( {{x^4} + {x^4}} \right) + \left( { - 5{x^3} - 2{x^3}} \right) + \left( {{x^2} + {x^2}} \right) + \left( {5x - 5x} \right) + \left( { - \dfrac{1}{3} - \dfrac{2}{3}} \right)\\ = 2{x^4} - 7{x^3} + 2{x^2} - 1\end{array}\)

\(\begin{array}{l}A\left( x \right) - B\left( x \right)\\ = \left( {{x^4} - 5{x^3} + {x^2} + 5x - \dfrac{1}{3}} \right) - \left( {{x^4} - 2{x^3} + {x^2} - 5x - \dfrac{2}{3}} \right)\\ = {x^4} - 5{x^3} + {x^2} + 5x - \dfrac{1}{3} - {x^4} + 2{x^3} - {x^2} + 5x + \dfrac{2}{3}\\ = \left( {{x^4} - {x^4}} \right) + \left( { - 5{x^3} + 2{x^3}} \right) + \left( {{x^2} - {x^2}} \right) + \left( {5x + 5x} \right) + \left( { - \dfrac{1}{3} + \dfrac{2}{3}} \right)\\ =  - 3{x^3} + 10x + \dfrac{1}{3}\end{array}\)