Cho các đa thức \(A\left( x \right) = 2{x^3} – 2{x^2} + x – 4;B\left( x \right) = 3{x^3} – 2x + 3;C\left( x \right) = – {x^3} + 1\). Hãy tính:
a)\(A\left( x \right) + B\left( x \right) + C\left( x \right);\)
b) \(A\left( x \right) – B\left( x \right) – C\left( x \right).\)
\(\begin{array}{l}A + B + C = A + \left( {B + C} \right)\\A – B – C = A – \left( {B + C} \right)\end{array}\)
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Tính B + C
a)
\(\begin{array}{l}A\left( x \right) + B\left( x \right) + C\left( x \right)\\ = A\left( x \right) + \left[ {B\left( x \right) + C\left( x \right)} \right]\\ = 2{x^3} – 2{x^2} + x – 4 + \left( {3{x^3} – 2x + 3 – {x^3} + 1} \right)\\ = 2{x^3} – 2{x^2} + x – 4 + 2{x^3} – 2x + 4\\ = \left( {2{x^3} + 2{x^3}} \right) – 2{x^2} + \left( {x – 2x} \right) + \left( { – 4 + 4} \right)\\ = 4{x^3} – 2{x^2} – x\end{array}\)
b)
\(\begin{array}{l}A\left( x \right) – B\left( x \right) – C\left( x \right)\\ = A\left( x \right) – \left[ {B\left( x \right) + C\left( x \right)} \right]\\ = 2{x^3} – 2{x^2} + x – 4 – \left( {3{x^3} – 2x + 3 – {x^3} + 1} \right)\\ = 2{x^3} – 2{x^2} + x – 4 – 2{x^3} + 2x – 4\\ = \left( {2{x^3} – 2{x^3}} \right) – 2{x^2} + \left( {x + 2x} \right) + \left( { – 4 – 4} \right)\\ = – 2{x^2} + 3x – 8\end{array}\)