Câu 1:
a, * A=(x−2)3−x2(x−4)+8A=(x-2)3-x2(x-4)+8 (x≠3)(x≠3)
A=x3−6x2+12x−8−x3+4x2+8A=x3-6x2+12x-8-x3+4x2+8
A=−2x2+12x=−2x(x−6)A=-2x2+12x=-2x(x-6)
* B=(x2−6x+9):(x−3)−x(x+7)−9B=(x2-6x+9):(x-3)-x(x+7)-9 (x≠3)(x≠3)
B=(x−3)2:(x−3)−x2−7x−9B=(x-3)2:(x-3)-x2-7x-9
B=x−3−x2−7x−9B=x-3-x2-7x-9
B=−x2−6x−12B=-x2-6x-12
b, Với x=−1x=-1 (tmđk)
Thay vào AA ta được: A=−2.(−1).(−1−6)=−14A=-2.(-1).(-1-6)=-14
Vậy, với x=−1x=-1 thì A=−14A=-14
c, Có C=A+BC=A+B
C=−2x(x−6)+(−x2−6x−12)C=-2x(x-6)+(-x2-6x-12)
C=−2x2+12x−x2−6x−12C=-2x2+12x-x2-6x-12
C=−3x2+6x−12C=-3x2+6x-12
C=−3(x2−2x+1)−9C=-3(x2-2x+1)-9
C=−3(x−1)2−9C=-3(x-1)2-9
Có (x−1)2≥0∀x∈R(x-1)2≥0∀x∈R ⇔ −3(x−1)2≤0-3(x-1)2≤0 ⇔ −3(x−1)2−9≤−9<0-3(x-1)2-9≤-9<0
Hay C<0∀x∈RC<0∀x∈R
Vậy C<0∀x∈R