\(A=x^2-x-1\)
\(A=x^2-x+\dfrac{1}{4}-\dfrac{5}{4}\)
\(A=\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge\dfrac{5}{4}\)
Dấu "=" xảy ra khi:
\(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
\(B=3x-x^2+2\)
\(B=-x^2+3x+2\)
\(B=-x^2+3x-\dfrac{9}{4}+\dfrac{17}{4}\)
\(B=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{17}{4}\)
\(B=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{17}{4}\le\dfrac{17}{4}\)
Dấu "=" xảy ra khi:
\(-\left(x-\dfrac{3}{2}\right)^2=0\Rightarrow x=\dfrac{3}{2}\)
a,Câu a phải tìm nhỉ nhất nhé
\(A=x^2-x-1\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\)
Với mọi x thì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)
=>\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
Hay \(A\ge-\dfrac{5}{4}\) với mọi x
Để \(A=-\dfrac{5}{4}\) thì \(\left(x-\dfrac{1}{2}\right)^2=0\)
=>\(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
Vậy...
b/\(B=3x-x^2+2\)
\(=-\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{17}{4}\)
\(=\dfrac{17}{4}-\left(x+\dfrac{3}{2}\right)^2\)
Với mọi x thì \(\left(x+\dfrac{3}{2}\right)^2\ge0\)
=>\(\dfrac{17}{4}-\left(x+\dfrac{3}{2}\right)^2\le\dfrac{17}{4}\)
Hay \(B\le\dfrac{17}{4}\) với mọi x
Để \(B=\dfrac{17}{4}\) thì
\(\left(x+\dfrac{3}{2}\right)^2=0\Rightarrow x+\dfrac{3}{2}=0\Rightarrow x=-\dfrac{3}{2}\)
Vậy...
a) ta có:
A= x2 - x -1
= (x2 - 2.x.1/2 + 1/4) -5/4
= (x - 1/2)2 - 5/4
vì (x - 1/2)2 \(\ge\) 0 \(\)
\(\Rightarrow\)(x-1/2)2 -5/4 \(\ge\)-5/4
vậy GTNN của A bằng -5/4
b.B = 3x - x2 +2
ta có: B= 3x - x2 + 2
= -x2 +3x +2
= -(x2 -3x) +2
= -(x2 -2.x.3/2 +9/4) +17/4
= -(x - 3/2)2 +17/4
vì: -(x - 3/2)2 \(\le\) 0
\(\Rightarrow\) -(x-3/2)2 +17/4 \(\le\) 17/4
vậy GTLN của biểu thức B = 17/4
