Rút gọn các biểu thức sau: B = (x – 2)(x^2 – 5x + 7) – (x^2 – 3x)(x – 4) – 5(x – 2).
Giải thích
Đặt E = (x – 2)(x2 – 5x + 7) và F = (x2 – 3x)(x – 4), ta có B = E – F – 5(x – 2).
Trước hết ta tính:
E = (x – 2)(x2 – 5x + 7) = x(x2 – 5x + 7) – 2(x2 – 5x + 7)
= (x3 – 5x2 + 7x) – (2x2 – 10x + 14)
= x3 + (– 5x2 – 2x2) + (7x + 10x) – 14
= x3 – 7x2 + 17x – 14
F = (x2 – 3x)(x – 4) = x2(x – 4) – 3x(x – 4)
= (x3 – 4x2) – (3x2 – 12x)
= x3 + (– 4x2 – 3x2) + 12x
= x3 – 7x2 + 12x
Cuối cùng ta được:
B = E – F – 5(x – 2) = (x3 – 7x2 + 17x – 14) – (x3 – 7x2 + 12x) – 5(x – 2)
= x3 – 7x2 + 17x – 14 – x3 + 7x2 – 12x – 5x + 10
= (x3 – x3) + (– 7x2 + 7x2) + (17x – 12x – 5x) + (10 – 14)
= – 4.