Question

cau1:

cho da thuc A(x)=ax²+bx+c.Biết 5a+b+2c=0.Chứng minh A(2).A(-1)_<0

câu2

 

Answer 1

Ta có:

\(A\left(2\right)=a.2^2+b.2+c=4a+2b+c\left(1\right)\)

\(A\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\left(2\right)\)

Lấy (1)+(2),ta đc:

\(A\left(2\right)+A\left(-1\right)=\left(4a+2b+c\right)+\left(a-b+c\right)=\left(4a+a\right)+\left(2b-b\right)+\left(c-c\right)\)

\(=5a+b+2c=0\)

=>\(A\left(2\right)=-A\left(-1\right)\)

=>\(A\left(2\right).A\left(-1\right)=-A\left(-1\right).A\left(-1\right)=-\left[A\left(-1\right)\right]^2\le0\) (đpcm)

Answer 2

A(2)=a.2²+b.2+c=4a+2b+c

A(-1)=a.(-1)²+(-1).b+c=a-b+c

=> A(2) + A(-1) = 5a+b+2c=0 (theo gia thiet)

=> A(2) = -A(-1)

=> A(2).A(-1) = -A(-1).A(-1)=- <A(-1)>² < hoac =0

Dấu = xảy ra khi a=b=c=0