Ta có:
\(B=-5x^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9x^2\)
\(=\left(2x-1\right)^2-\left(3x\right)^2\)
\(=\left(2x-1+3x\right)\left(2x-1-3x\right)\)
\(=-\left(x+1\right)\left(5x-1\right)\)
\(B=-5x^2-4x+1\)
\(B=-5\left(x^2+\frac{4}{5}x-\frac{1}{5}\right)\)
\(B=-5\left[x^2+2.x.\frac{2}{5}+\left(\frac{2}{5}\right)^2-\frac{9}{25}\right]\)
\(B=-5\left(x+\frac{2}{5}\right)^2+5.\frac{9}{25}\)
\(B=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\)
Ta có: \(\left(x+\frac{2}{5}\right)^2\ge0\forall x\)
\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2\le0\forall x\)
\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\forall x\)
\(B=\frac{9}{5}\Leftrightarrow-5.\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)
Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)
Tham khảo nhé~
\(B=-5x^2-4x+1\)
\(=-5x^2-4x-\frac{4}{5}+\frac{9}{5}\)
\(=\left(-5x^2-4x-\frac{4}{5}\right)+\frac{9}{5}\)
\(=-5\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{9}{5}\)
\(=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)(vì \(-5\left(x+\frac{2}{5}\right)^2\le0\forall x\))
Dấu "=" xảy ra \(\Leftrightarrow-5\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x=-\frac{2}{5}\)
Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)
\(B=-5x^2-4x+1\)
\(=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)
Dấu \(=\)khi \(x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)
Vậy GTLN của \(B\)là \(\frac{9}{5}\)đạt khi \(x=-\frac{2}{5}\).
