a, \(A=-7x^2+2x+8\)
\(A=-\left(7x^2-2x-8\right)\)
\(A=-\left(7x^2-x-x+\dfrac{1}{7}+\dfrac{55}{7}\right)\)
\(A=-\left[x\left(7x-1\right)-\dfrac{1}{7}\left(7x-1\right)+\dfrac{55}{7}\right]\)
\(A=-\left[\dfrac{1}{7}\left(7x-1\right)^2+\dfrac{55}{7}\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\dfrac{1}{7}\left(7x-1\right)^2\ge0\Rightarrow\dfrac{1}{7}\left(7x-1\right)^2+\dfrac{55}{7}\ge\dfrac{55}{7}\)
\(\Rightarrow-\left[\dfrac{1}{7}\left(7x-1\right)^2+\dfrac{55}{7}\right]\le-\dfrac{55}{7}\)
Hay \(A\le-\dfrac{55}{7}\) với mọi giá trị của x.
Để \(A=-\dfrac{55}{7}\) thì \(-\left[\dfrac{1}{7}\left(7x-1\right)^2+\dfrac{55}{7}\right]=-\dfrac{55}{7}\)
\(\Rightarrow\left(7x-1\right)^2=0\Rightarrow x=\dfrac{1}{7}\)
Vậy.....................
b, \(B=3x^2-2x+8\)
\(B=3x^2-x-x+\dfrac{1}{3}+\dfrac{23}{3}\)
\(B=x\left(3x-1\right)-\dfrac{1}{3}\left(3x-1\right)+\dfrac{23}{3}\)
\(B=\dfrac{1}{3}\left(3x-1\right)^2+\dfrac{23}{3}\)
Với mọi giá trị của \(x\in R\) ta có:
\(\dfrac{1}{3}\left(3x-1\right)^2\ge0\Rightarrow\dfrac{1}{3}\left(3x-1\right)^2+\dfrac{23}{3}\ge\dfrac{23}{3}\)
Hay \(A\ge\dfrac{23}{3}\) với mọi giá trị của x.
Để \(A=\dfrac{23}{3}\) thì \(\dfrac{1}{3}\left(3x-1\right)^2+\dfrac{23}{3}=\dfrac{23}{3}\)
\(\Rightarrow\left(3x-1\right)^2=0\Rightarrow x=\dfrac{1}{3}\)
Vậy..........
Chúc bạn học tốt!!!
a, Sửa đề:
\(A=-7x^2+2x+8=-7\left(x^2-\dfrac{2}{7}x+\dfrac{1}{49}\right)+\dfrac{57}{7}\)\(=-7\left(x-\dfrac{1}{7}\right)^2+\dfrac{57}{7}\le\dfrac{57}{7}\forall x\)
Vậy Max A = \(\dfrac{57}{7}\)khi \(x-\dfrac{1}{7}=0\Rightarrow x=\dfrac{1}{7}\)
\(b,B=3x^2-2x+8=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)+\dfrac{23}{3}\)\(=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{23}{3}\ge\dfrac{23}{3}\forall x\)
Vậy Min B = \(\dfrac{23}{3}\) khi \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
