\(a,P\left(x\right)=-x^2-7x+1=-\left(x^2+7+\dfrac{49}{4}\right)+\dfrac{53}{4}=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{53}{4}\le\dfrac{53}{4}\)Vậy \(Max_{P\left(x\right)}=\dfrac{53}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=-\dfrac{7}{2}\)
\(b,Q\left(x\right)=-2x^2+x-8=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{63}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{-63}{8}\le\dfrac{-63}{8}\)Vậy \(Max_{Q\left(x\right)}=\dfrac{-63}{8}\)khi \(x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{4}\)
\(P=-x^2-7x+1\)
\(-P=x^2+7x-1\)
\(=\left(x^2+2.\dfrac{7}{2}x+\dfrac{49}{4}\right)-\dfrac{53}{4}\)
\(=\left(x+\dfrac{7}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{53}{4}\)
\(=>MIN_P=-\dfrac{53}{4}=>MAX_P=\dfrac{53}{4}\Leftrightarrow x=\dfrac{7}{2}\)
\(b,Q=-2x^2+x-8\)
\(-Q=2x^2-x+8\)
\(=2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{63}{8}\)
\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{63}{8}\ge\dfrac{63}{8}\)
\(=>MIN_{-Q}=\dfrac{63}{8}=>MAX_Q=-\dfrac{63}{8}\Leftrightarrow x=\dfrac{1}{4}\)
a, \(P\left(x\right)=-x^2-7x+1=-\left(x^2+7x-1\right)\)
\(=-\left(x^2+\dfrac{7}{2}.x.2+\dfrac{49}{4}-\dfrac{53}{4}\right)\)
\(=-\left[\left(x+\dfrac{7}{2}\right)^2-\dfrac{53}{4}\right]\)
\(=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{53}{4}\)
Ta có: \(-\left(x+\dfrac{7}{2}\right)^2\le0\Leftrightarrow P\left(x\right)=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{53}{4}\le\dfrac{53}{4}\)
Dấu " = " khi \(-\left(x+\dfrac{7}{2}\right)^2=0\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(MAX_{P\left(x\right)}=\dfrac{53}{4}\) khi \(x=\dfrac{-7}{2}\)
b, \(Q\left(x\right)=-2x^2+x-8\)
\(=-2\left(x^2-2.x.\dfrac{1}{4}+4\right)\)
\(=-2\left(x^2-2x.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{63}{16}\right)\)
\(=-2\left[\left(x-\dfrac{1}{4}\right)^2+\dfrac{63}{16}\right]\)
\(=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{63}{8}\le\dfrac{-63}{8}\)
Dấu " = " khi \(-2\left(x-\dfrac{1}{4}\right)^2=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(MAX_{Q\left(x\right)}=\dfrac{-63}{8}\) khi \(x=\dfrac{1}{4}\)
phải tìm GTNN k bn, nếu z thì mk làm cái này nha còn nếu là GTLN thì mk bó tay :3
a) Ta có: \(P\left(x\right)=-x^2-7x+1=-\left(x^2+7x-1\right)\)
\(=-\left[\left(x+\dfrac{7}{2}\right)^2-\dfrac{49}{4}-1\right]\)
\(=-\left[\left(x+\dfrac{7}{2}\right)^2-\dfrac{53}{4}\right]=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{53}{4}\ge\dfrac{53}{4}\)
Vậy \(MIN_{P\left(x\right)}=\dfrac{53}{4}\) tại \(x=-\dfrac{7}{2}\)
