Question

 Cho đa thức: H(x)= ax²+bx+c.  Biết 5a-3b+2c=0, hãy chứng tỏ rằng: H(-1).H(-2) \(\le\)0

Answer 1

Tính H(-1) = a.(-1)² + b.(-1) + c = a - b + c

H(-2) = a.(-2)² + b.(-2) + c = 4a - 2b + c

=> H(-1) + H(-2) = 5a - 3b + 2c = 0 

=> H(-1) = - H(-2)

=> H(-1) . H(-2) = [- H(-2)].h(-2) = - H²(-2) \(\le\) 0 Vì H²(-2) \(\ge\) 0

=> ĐPCM

Answer 2

Ta có \(H\left(-1\right)=a-b+c;H\left(-2\right)=4a-2b+c\)

\(\Rightarrow H\left(-1\right)+H\left(-2\right)=a-b+c+4a-2b+c=5a-3b+2c=0\left(1\right)\)

\(\Rightarrow H\left(-1\right)=-H\left(-2\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow H\left(-1\right)\cdot H\left(-2\right)=-H\left(-2\right)\cdot H\left(-2\right)=-\left[H\left(-2\right)\right]^2=\le0\)