Question

Cho f(x) = ax² + bx + c vs a,b,c là số hữu tỉ . CT:

f(-2) . f(3) \(\le\) 0 biết 13a + b + 2c = 0

Answer 1

\(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)

\(\Rightarrow f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)

\(\Rightarrow f\left(-2\right)+f\left(3\right)=4a-2b+c+9a+3b+c=13a+b+2c=0\)

\(\Rightarrow f\left(-2\right)+f\left(3\right)=0\Rightarrow f\left(-2\right)=-f\left(3\right)\)

Xét \(f\left(-2\right).f\left(3\right)=\left[-f\left(3\right)\right].f\left(3\right)=-\left[f\left(3\right)\right]^2\le0\)

Vậy \(f\left(-2\right).f\left(3\right)\le0\)

Answer 2

mình không hiểu, sao f(−2).f(3)=[−f(3)].f(3)=−[f(3)]2?