a,Đặt \(A=-3x^2-6x-4=-3\left(x^2+2x+\dfrac{4}{3}\right)\)
\(=-3\left(x^2+2x+1+\dfrac{1}{3}\right)\)
\(=-3\left[\left(x+1\right)^2+\dfrac{1}{3}\right]\)
\(=-3\left(x+1\right)^2-1\le-1\)
Dấu " = " khi \(-3\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Vậy \(MAX_A=-1\) khi x = -1
b, Đặt \(B=-5x^2+8x=-5\left(x^2-\dfrac{4}{5}x.2+\dfrac{16}{25}-\dfrac{16}{25}\right)\)
\(=-5\left(x-\dfrac{4}{5}\right)^2+\dfrac{16}{5}\le\dfrac{16}{5}\)
Dấu " = " khi \(-5\left(x-\dfrac{4}{5}\right)^2=0\Leftrightarrow x=\dfrac{4}{5}\)
Vậy \(MAX_B=\dfrac{16}{5}\) khi \(x=\dfrac{4}{5}\)
a, \(C=-3x^2-6x-4\)
\(=>-C=3x^2+6x+4\)
\(=3\left(x^2+2x+1\right)+1\)
\(=3\left(x+1\right)^2+1\ge1\)
\(=>MIN_{-C}=1=>MAX_C=-1\Leftrightarrow x=-1\)
\(b,T=-5x^2+8x\)
\(-T=5x^2-8x=5\left(x^2-2.\dfrac{4}{5}x+\dfrac{16}{25}\right)-\dfrac{16}{5}\)
\(=\left(x-\dfrac{4}{5}\right)^2-\dfrac{16}{5}\ge\dfrac{-16}{5}\)
\(=>MIN_{-T}=\dfrac{-16}{5}=>MAX_T=\dfrac{16}{5}\Leftrightarrow x=\dfrac{4}{5}\)
a) Đặt \(A=-3x^2-6x-4\)
\(=-\left(3x^2+6x+4\right)\)
\(=-\left(3x^2+2\cdot x\cdot3+3^2-9+4\right)\)
\(=-\left[\left(x+3\right)^2-5\right]\)
\(=-\left(x+3\right)^2+5\le5\)
Dấu "=" xảy ra khi \(x=-3\)
Vậy \(MIN_A=5\) tại \(x=-3\)
b) Đặt \(B=-5x^2+8x\)
\(=-\left(5x^2-8x\right)\)
\(=-\left(5x^2-2\cdot x\cdot4+4^2-16+0\right)\)
\(=-\left[\left(x-4\right)-16\right]\)
\(=-\left(x-4x\right)+16\le16\)
Dấu "=" xảy ra khi \(x=4\)
Vậy \(MIN_B=16\) tại \(x=4\)
