\(A=4x^2+2x-3\)
\(A=4x^2+2+\dfrac{1}{4}-\dfrac{13}{4}\)
\(A=\left(2x+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\)
Vậy \(Min_A=-\dfrac{13}{4}\)
Dấu = xảy ra khi \(\left(2x+\dfrac{1}{2}\right)^2=0\Rightarrow x=-\dfrac{1}{4}\)
\(B=x^2-5x+1\)
\(B=x^2-5x+\dfrac{5}{2}-\dfrac{3}{2}\)
\(B=\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{2}\)
Vậy Min B \(=-\dfrac{3}{2}\)
Dấu = xảy ra khi \(\left(x-\dfrac{5}{2}\right)^2=0\Rightarrow x=\dfrac{5}{2}\)
A = 4x2 + 2x - 3
= 4x2 + 2x + \(\dfrac{1}{4}-\dfrac{13}{4}\)
= (2x+\(\dfrac{1}{2}\))2 - \(\dfrac{13}{4}\) \(\ge\) \(-\dfrac{13}{4}\)
=> MinA = \(-\dfrac{13}{4}\)
Dấu "=" xảy ra khi x - \(-\dfrac{1}{4}\)
B = x2 - 5x + 1
= x2 - 5x + \(\dfrac{25}{4}-\dfrac{21}{4}\)
= (x-\(\dfrac{5}{2}\))2 - \(\dfrac{21}{4}\ge-\dfrac{21}{4}\)
=> MinB = \(-\dfrac{21}{4}\)
Dấu "=" xảy ra khi x = \(\dfrac{5}{2}\)
\(A=4x^2+2x-3\)
\(A=4x^2+2x+\dfrac{1}{4}-\dfrac{13}{4}\)
\(A=\left(2x+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\)
Dấu "=" xảy ra khi:
\(2x=-\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{4}\)
\(B=x^2-5x+1\)
\(B=x^2-5x+\dfrac{25}{4}-\dfrac{29}{4}\)
\(B=\left(x+\dfrac{5}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\)
Dấu "=" xảy ra khi:
\(x=-\dfrac{5}{2}\)
