Bài 1: Cho \(P(x) = – 3{{\rm{x}}^2} + 3{\rm{x}} – 4{{\rm{x}}^3} + 5 – 2{{\rm{x}}^4};\)\(\;Q(x) = 5{{\rm{x}}^4} + 9{{\rm{x}}^2} + 4{{\rm{x}}^3} – 6{\rm{x}} – 12\).
a) Sắp xếp theo lũy thừa giảm dần của biến của P(x) và Q(x).
b) Tính \(P(x) + Q(x)\) và \(P(x) – Q(x)\).
Bài 2: Cho \(A(x) = {x^3} – 3{{\rm{x}}^2} + 3{\rm{x}} – 1;\)\(\;B(x) = {{\rm{x}}^3} + 3{{\rm{x}}^2} + 3{\rm{x}} + 1;\)\(\;C(x) = 2{{\rm{x}}^2} + 3{\rm{x}} + 2\).
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Tính \(A(x) – B(x) + C(x)\).
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Bài 1: \(a){\rm{ }}P(x) = – 2{{\rm{x}}^4} – 4{{\rm{x}}^3} – 3{{\rm{x}}^2} + 3{\rm{x}} + 5.\)
\({\rm{ }}Q(x) = 5{{\rm{x}}^4} + 4{{\rm{x}}^3} + 9{{\rm{x}}^2} – 6{\rm{x}} – 12.\)
b) \(P(x) + Q(x) = 3{{\rm{x}}^4} + 6{{\rm{x}}^2} – 3{\rm{x}} – 7.\)
\(\eqalign{ P(x) – Q(x) &= – 2{{\rm{x}}^4} – 4{{\rm{x}}^3} – 3{{\rm{x}}^2} + 3{\rm{x}} + 5 – (5{{\rm{x}}^4} + 4{{\rm{x}}^3} + 9{{\rm{x}}^2} – 6{\rm{x}} – 12) \cr & {\rm{ }} = – 2{{\rm{x}}^4} – 4{{\rm{x}}^3} – 3{{\rm{x}}^2} + 3{\rm{x}} + 5 – 5{{\rm{x}}^4} – 4{{\rm{x}}^3} – 9{{\rm{x}}^2} + 6{\rm{x + }}12 \cr & {\rm{ }} = – 7{{\rm{x}}^4} – 8{{\rm{x}}^3} – 12{{\rm{x}}^2} + 9{\rm{x}} + 17. \cr} \)
Bài 2: Ta có:
\(\eqalign{ A(x) – B(x) + C(x) &= ({x^3} – 3{{\rm{x}}^2} + 3{\rm{x}} – {\rm{1)}} – {\rm{(}}{{\rm{x}}^3} + 3{{\rm{x}}^2} + 3{\rm{x}} + 1) + (2{{\rm{x}}^2} + 3{\rm{x}} + 2) \cr & = {x^3} – 3{{\rm{x}}^2} + 3{\rm{x}} – {\rm{1}} – {{\rm{x}}^3} – 3{{\rm{x}}^2} – 3{\rm{x – }}1 + 2{{\rm{x}}^2} + 3{\rm{x}} + 2 \cr & = – 4{{\rm{x}}^2} + 3{\rm{x}}{\rm{.}} \cr} \)
