a) \(A=2x-x^2\)
\(=-\left(x^2-2x\right)=-\left[\left(x^2-2.x.1-1^2\right)+1\right]\)
\(=-\left[\left(x-1\right)^2+1\right]=-\left(x-1\right)^2-1\)
\(=1+\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+1\ge1\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(MIN_A=1\Leftrightarrow x=1.\)
b) \(B=19-6x-9x^2\)
\(=-\left(9x^2+6x-19\right)=-\left[9\left(x^2+\dfrac{6}{9}x-\dfrac{19}{9}\right)\right]\)
\(=-\left[9\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}x-\dfrac{19}{9}\right)\right]=-\left[9\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}x+\dfrac{1}{9}-\dfrac{20}{9}\right)\right]\)
\(=-\left[9\left(x+\dfrac{1}{3}\right)^2-20\right]\)
\(=-9\left(x+\dfrac{1}{3}\right)^2+20=20-9\left(x+\dfrac{1}{3}\right)^2\)
Do \(-9\left(x+\dfrac{1}{3}\right)^2\le0\Rightarrow20-9\left(x+\dfrac{1}{3}\right)^2\le20\)
Dấu "=" xảy ra khi \(x=\dfrac{-1}{3}\)
Vậy \(MAX_B=20\Leftrightarrow x=-\dfrac{1}{3}.\)
Câu b theo cách tách của mk thì là tìm GTLN thì đúng hơn.
Dấu của hạng tử bậc là dấu âm nên chỉ tìm được giá trị lớn nhất thôi nhé.
\(\text{a) }A=2x-x^2\\ A=2x-x^2+1-1\\ A=1-\left(x^2-2x+1\right)\\ A=1-\left(x-1\right)^2\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow A=1-\left(x-1\right)^2\le1\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ Vậy\text{ }Max_A=1\text{ }khi\text{ }x=1\)
\(\text{b) }B=19-6x-9x^2\\ B=20-1-6x-9x^2\\ B=20-\left(1+6x+9x^2\right)\\ B=20-\left(1+3x\right)^2\\ Do\text{ }\left(1+3x\right)^2\ge0\forall x\\ \Rightarrow B=20-\left(1+3x\right)^2\le20\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(1+3x\right)^2=0\\ \Leftrightarrow1+3x=0\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\\ Vậy\text{ }Max_B=20\text{ }khi\text{ }x=-\dfrac{1}{3}\)
