Câu hỏi của Nguyễn Thùy Linh

Question

cho đa thức: $f\left(x\right)=ax^2+bx+c$ biết $29a+2c=3b$ chứng minh rằng $f\left(2\right)\times f\left(-5\right)\le0$

Đoàn Đức Hà · Accepted Answer

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

$f\left(-5\right)=a.\left(-5\right)^2+b.\left(-5\right)+c=25a-5b+c$

$f\left(2\right)+f\left(5\right)=4a+2b+c+25a-5b+c=29a-3b+2c$

$=\left(29a+2c\right)-3b=3b-3b=0$

$\Leftrightarrow f\left(2\right)=-f\left(-5\right)$

$\Leftrightarrow f\left(2\right)f\left(-5\right)\le0$.

Câu trả lời: $f\left(2\right)=a.2^2+b.2+c=4a+2b+c$

$f\left(-5\right)=a.\left(-5\right)^2+b.\left(-5\right)+c=25a-5b+c$

$f\left(2\right)+f\left(5\right)=4a+2b+c+25a-5b+c=29a-3b+2c$

$=\left(29a+2c\right)-3b=3b-3b=0$

$\Leftrightarrow f\left(2\right)=-f\left(-5\right)$

$\Leftrightarrow f\left(2\right)f\left(-5\right)\le0$.