\(D=-3x^2+4x+3\)
\(\Leftrightarrow D=-3\left(x^2-\dfrac{4x}{3}-1\right)\)
\(\Leftrightarrow D=-3\left(x^2-2.x.\dfrac{4}{6}+\dfrac{16}{36}-\dfrac{13}{9}\right)\)
\(\Leftrightarrow D=-3\left[\left(x-\dfrac{4}{6}\right)^2-\dfrac{13}{9}\right]\)
\(\Leftrightarrow D=-3\left(x-\dfrac{4}{6}\right)^2+\dfrac{13}{3}\le\dfrac{13}{3},\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{4}{6}\)
Vậy Max D = \(\dfrac{13}{3}\Leftrightarrow x=\dfrac{4}{6}\)
\(A=x^2-x+1\)
\(\Leftrightarrow A=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của A là \(\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
\(B=\left(2-x\right)\left(4+x\right)+10\)
\(\Leftrightarrow B=8+2x-4x-x^2+10\)
\(\Leftrightarrow B=-x^2-2x+18\)
\(\Leftrightarrow B=-\left(x^2+2x-18\right)\)
\(\Leftrightarrow B=-\left(x^2+2x+1-19\right)\)
\(\Leftrightarrow B=-\left[\left(x+1\right)^2-19\right]\)
\(\Leftrightarrow B=-\left(x+1\right)^2+19\le19,\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy Max B = 19 \(\Leftrightarrow x=-1\)
câu C: Câu hỏi của trần như - Toán lớp 8 - Học toán với OnlineMath
