Bài 1:
\(A = 2{{\rm{a}}^2} + ab - {b^2} + {a^2} - {b^2} + ab \)\(\;= 3{{\rm{a}}^2} + 2{\rm{a}}b - 2{b^2};\)
\(B = 3{{\rm{a}}^2} + {b^2} - ab + {a^2} \)\(\;= 4{{\rm{a}}^2} - ab + {b^2}\).
a) \(A + B = (3{{\rm{a}}^2} + 2{\rm{a}}b - 2{b^2}) \)\(\;+ (4{{\rm{a}}^2} - ab + {b^2}) \)\(\;= 7{{\rm{a}}^2} + ab - {b^2}.\)
b) \(A - B = (3{{\rm{a}}^2} + 2{\rm{a}}b - 2{b^2}) - (4{{\rm{a}}^2} - ab + {b^2})\)
\(\eqalign{ & = 3{a^2} + 2ab - 2{b^2} - 4{a^2} + ab - {b^2} \cr & = - {a^2} + 3ab - 3{b^2}. \cr} \)
Bài 2: Ta có:
\(\eqalign{ & f(x) + g(x) - h(x) \cr & = (2{x^5} - 3{x^4} + 5{x^2} - 6) + (3{x^5} - 2{x^4} + 3{x^3} + 3) - ({x^5} + 2{x^3} - 7x + 4) \cr & = 2{x^5} - 3{x^4} + 5{x^2} - 6 + 3{x^5} - 2{x^4} + 3{x^3} + 3 - {x^5} - 2{x^3} + 7x - 4 \cr & = 4{x^5} - 5{x^4} + {x^3} + 5{x^2} + 7x - 7. \cr} \)
Thay \(x = - 1\) vào biểu thức trên, ta được:
\(\eqalign{ & f( - 1) + g( - 1) - h( - 1) \cr & = 4{( - 1)^5} - 5{( - 1)^4} + {( - 1)^3} + 5{( - 1)^2} + 7( - 1) - 7 \cr & = - 4 - 5 - 1 + 5 - 7 - 7 = - 19. \cr} \)
Bài 3:
a) Ta có: \(M(1) = - 3 \)\(\Rightarrow {1^2} - 2m.1 + m - 2 = - 3\)\( \Rightarrow - m - 1 = - 3 \Rightarrow m = 2.\)
b) Khi \(m = 2\), ta có \(M(x) = {x^2} - 4{\rm{x}}.\)
\(M(x) = 0 \Rightarrow {x^2} - 4{\rm{x}} = 0\)\(\; \Rightarrow x(x - 4) = 0\)
