Question

Bài 1; Rút gọn

a, \(x^5-x^4-1\)

b, \(\left(x^2+x\right)^2+4x^2+4x-12\)

c, \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Answer 1

a, \(x^5-x^4-1\\ =x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\\ =x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\\ =\left(x^2-x+1\right)\left(x^3-x-1\right)\)

b, \(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x\right)^2+4\left(x^2+x\right)-12\\ =\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12\\ =\left(x^2+x-2\right)\left(x^2+x\right)+6\left(x^2+x-2\right)\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c. Đặt x²+x+1=t

Khi đó ta có t(t+1)-12=t²+t-12=t²-3t+4t-12=t(t-3)+4(t-3)=(t-3)(t+4)

Thay t=x²+x+1 t đc (x²+x+1-3)(x²+x+1+4)=(x²+x-2)(x²+x+5)

=(x-1)(x+2)(x²+x+5)