Bài 1:
a) \(P = - 2{{\rm{x}}^3} + x{y^2} + {x^2}y - 1 - {x^2}y + x{y^2} - 3{{\rm{x}}^3} \)
\(\;\;\;\;= - 5{{\rm{x}}^3} + 2{\rm{x}}{y^2} - 1.\)
b) \(Q = 4{{\rm{a}}^2} - 2{\rm{a}}b - {b^2} + {a^2} - {b^2} + 2{\rm{a}}b + 3{{\rm{a}}^2} + {b^2} - ab \)
\(\;\;\;\;= 8{{\rm{a}}^2} - ab - {b^2}\).
Bài 2: Ta có
\(\eqalign{ & A - M = B \cr&\Leftrightarrow 2{{\rm{x}}^2} - 6{\rm{x}} + 5 - M = {x^2} + 6x - 8 \cr & {\rm{ }} \Rightarrow M = (2{{\rm{x}}^2} - 6{\rm{x}} + 5) - ({x^2} + 6x - 8) \cr & {\rm{ }} \;\;\;\;\;\;\;\;\;\;= 2{x^2} - 6x + 5 - {x^2} - 6x + 8 \cr & \;\;\;\;\;\;\;\;\;\;\; = {x^2} - 12x + 13. \cr} \)
Bài 3: Ta có:
\(\eqalign{ K + L &= (3{x^2} + 4xy - 2{y^2}) + ( - {x^2} + 3{y^2} - 4xy) \cr & = 3{x^2} + 4xy - 2{y^2} - {x^2} + 3{y^2} - 4xy \cr&= 2{{\rm{x}}^2} + {y^2}. \cr} \)
Do đó \(K + L = 2{{\rm{x}}^2} + {y^2} \ge 0\) với mọi x; y.
