\(3x^2+5x-6\)
\(=3x^2+5x+\dfrac{25}{12}-\dfrac{97}{12}\)
\(=\left(3x^2+5x+\dfrac{25}{12}\right)-\dfrac{97}{12}\)
\(=3\left(x^2+\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{97}{12}\)
\(=3\left[x^2+2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2\right]-\dfrac{97}{12}\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{97}{12}\ge\dfrac{-97}{12}\)
Vậy GTNN của đa thức trên \(=\dfrac{-97}{12}\) khi \(x+\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{-5}{6}\)
A=3x2+5x-6
=3x2+5x+\(\dfrac{25}{12}-\dfrac{97}{12}\)
=\(3.\left(x^2+2.\dfrac{5}{6}x+\dfrac{25}{36}\right)-\dfrac{97}{12}\)
=\(3.\left(x+\dfrac{5}{6}\right)^2-\dfrac{97}{12}\)
do \(\left(x+\dfrac{5}{6}\right)^2\ge0\forall x\)
=> \(3.\left(x+\dfrac{5}{6}\right)^2\ge0\)
=> \(3.\left(x+\dfrac{5}{6}\right)^2-\dfrac{97}{12}\ge-\dfrac{97}{12}\)
=> A \(\ge-\dfrac{97}{12}\)
=> GTNN A = \(-\dfrac{97}{12}\) khi \(x+\dfrac{5}{6}=0\)
=> x=\(-\dfrac{5}{6}\)
vâỵ GTNN A=\(-\dfrac{97}{12}khix=\dfrac{-5}{6}\)
