Bài 1: Thu gọn đa thức:
a) \(A = 5{{\rm{x}}^2} + 6{{\rm{x}}^3} + ({x^3} – {x^2}) – ( – 2{{\rm{x}}^3} + 4{{\rm{x}}^2});\)
b) \(B = 2{{\rm{a}}^2} – ({b^2} – 3{{\rm{a}}^2}) – {\rm{[}}5{{\rm{a}}^2} – 11{\rm{a}}b + 8{b^2} – ( – 2{b^2} – 7{{\rm{a}}^2} + 5{\rm{a}}b){\rm{]}}.\)
Bài 2: Cho \(K = {a^2} + ab – {b^2};\)\(\;M = 2{{\rm{a}}^2} + 3{\rm{a}}b – 5{b^2};\)\(\;L = – 4{{\rm{a}}^2} + 2{\rm{a}}b – 3{b^2}\).
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Tính \(K – M – L\).
Bài 3: Tìm đa thức P, biết: \(3{{\rm{x}}^2} + 3{{\rm{x}}^2}{y^2} – {x^3} – P = 3{{\rm{x}}^2} + 2{\rm{x}}y – 4{y^2}\).
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Bài 1: a) \(A = 5{{\rm{x}}^2} + 6{{\rm{x}}^3} + {x^3} – {x^2} + 2{{\rm{x}}^3} – 4{{\rm{x}}^2}\)\(\; = 9{{\rm{x}}^3}.\)
b) \(\eqalign{ B &= 2{{\rm{a}}^2} – {b^2} + 3{{\rm{a}}^2} – (5{{\rm{a}}^2} – 11{\rm{a}}b + 8{b^2} + 2{b^2} + 7{{\rm{a}}^2} – 5{\rm{a}}b) \cr & {\rm{ }} = 2{a^2} – {b^2} + 3{a^2} – 5{a^2} + 11ab – 8{b^2} – 2{b^2} – 7{a^2} + 5ab \cr & {\rm{ }} = – 7{a^2} – 11{b^2} + 16ab. \cr} \)
Bài 2: Ta có:
\(\eqalign{ K – M – L &= ({a^2} + ab – {b^2}) – (2{{\rm{a}}^2} + 3{\rm{a}}b – 5{b^2}) – ( – 4{{\rm{a}}^2} + 2{\rm{a}}b – 3{b^2}) \cr & {\rm{ }} = {a^2} + ab – {b^2} – 2{{\rm{a}}^2} – 3{\rm{a}}b + 5{b^2} + 4{{\rm{a}}^2} – 2{\rm{a}}b + 3{b^2} \cr & {\rm{ }} = 3{a^2}{\rm{ – 4a}}b + 7{b^2}. \cr} \)
Bài 3: Ta có:
\(\eqalign{ & 3{{\rm{x}}^2} + 3{{\rm{x}}^2}{y^2} – {x^3} – P = 3{{\rm{x}}^2} + 2{\rm{x}}y – 4{y^2} \cr & \Rightarrow P = 3{{\rm{x}}^2} + 3{{\rm{x}}^2}{y^2} – {x^3} – 3{{\rm{x}}^2} – 2{\rm{x}}y + 4{y^2} \cr & \Rightarrow P = 3{{\rm{x}}^2}{y^2} – {x^3} – 2{\rm{x}}y + 4{y^2}. \cr} \)
