a) (2x-1)(5x^2-4x+3) b) x (1-x) (2-2x)-(x-5) (x+4) tinh gia tri bieu thuc

(2x−1)(5$x^{2}$ −4x+3)(2x-1)(5$x^{2}$-4x+3)

=2x.(5$x^{2}$−4x+3)−1.(5$x^{2}$−4x+3)

=10$x^{3}$−8$x^{2}$+6x−5$x^{2}$+4x−3

=10$x^{3}$−(8$x^{2}$+5$x^{2}$)+(6x+4x)−3

=10$x^{3}$-13$x^{2}$+10x-3

b)

x(1−x)(2−2x)−(x+5)(x+4)x(1-x)(2-2x)-(x+5)(x+4)

=x(2−2x−2x+2$x^{2}$)−($x^{2}$+4x+5x+20)

=x(2$x^{2}$−4x+2)−($x^{2}$+9x+20)

=2$x^{3}$−4$x^{2}$+2x−$x^{2}$−9x−20

=2$x^{3}$−(4$x^{2}$+$x^{2}$)+(2x−9x)−20

=2$x^{3}$−5$x^{2}$−7x−20