Question

tìm giá trị lớn nhất

B= -5x²-4x- \(\dfrac{19}{5}\)

mk cần gấp ạ!!!

Answer 1

-Câu cuối thôi nha bạn :v

\(B=-5x^2-4x-\dfrac{19}{5}=-5\left(x^2+\dfrac{4}{5}x+\dfrac{19}{25}\right)=-5\left(x^2+2.\dfrac{2}{5}x+\dfrac{4}{25}+\dfrac{15}{25}\right)=-5\left(x+\dfrac{2}{5}\right)^2-\dfrac{15}{5}\le-3\)\(B_{max}=-3\Leftrightarrow x=\dfrac{-2}{5}\)