Môn Toán Lớp 8: 1 tính a(x-½)(x³+3x²-4x+7) b(x²-8)(x+1)(x+2)

Giải đáp:

 a)

(x-\frac{1}{2}).(x^{3}+3x^{2}-4x+7)

=x.(x^{3}+3x^{2}-4x+7)-\frac{1}{2}.(x^{3}+3x^{2}-4x+7)

=x^{4}+3x^{3}-4x^{2}+7x-\frac{1}{2}x^{3}-\frac{3}{2}x^{2}+2x-\frac{7}{2}

=x^{4}+(3x^{3}-\frac{1}{2}x^{3})-(4x^{2}+\frac{3}{2}x^{2})+(7x+2x)-\frac{7}{2}

=x^{4}+\frac{5}{2}x^{3}-\frac{11}{2}x^{2}+9x-\frac{7}{2}

b)

(x^{2}-8).(x+1).(x+2)

= (x^{2}-8).(x^{2}+2x+x+2)

=(x^{2}-8).(x^{2}+3x+2)

=x^{2}.(x^{2}+3x+2)-8.(x^{2}+3x+2)

=x^{4}+3x^{3}+2x^{2}-8x^{2}-24x-16

=x^{4}+3x^{3}-6x^2-24x-16