A=3x2+2x-5
=3x2+2x+\(\dfrac{1}{3}\) -\(\dfrac{16}{3}\)
=3\(\left(x^2+\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{16}{3}\)
=3\(\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\)
do \(\left(x+\dfrac{1}{3}\right)^2\ge0\forall x\)
=>\(3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\)
=>GTNNcủa A =-\(\dfrac{16}{3}\) khi
x+\(\dfrac{1}{3}=0\)
=>x=-\(\dfrac{1}{3}\)
vậy ...
\(A=3x^2+2x-5\)
\(=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\dfrac{1}{\sqrt{3}}+\left(\dfrac{1}{\sqrt{3}}\right)^2-\left(\dfrac{1}{\sqrt{3}}\right)^2-5\)
\(=\left[\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\dfrac{1}{\sqrt{3}}+\left(\dfrac{1}{\sqrt{3}}\right)^2\right]-\dfrac{1}{3}-5\)
\(=\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{16}{3}\)
Ta có :
\(\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\) với mọi x
Dấu = xảy ra khi
\(\left(\sqrt{3}x+\dfrac{1}{\sqrt{3}}\right)^2=0\)
\(\Leftrightarrow\sqrt{3}x+\dfrac{1}{\sqrt{3}}=0\)
\(\Leftrightarrow\sqrt{3}x=-\dfrac{1}{\sqrt{3}}\Leftrightarrow-\dfrac{1}{3}\)
Vậy \(Min_A=-\dfrac{16}{3}\Leftrightarrow x=-\dfrac{1}{3}\)
\(A=3x^2+2x-5\)
\(=3x^2-3x+5x-5\)
\(=\left(3x^2-3x\right)+\left(5x-5\right)\)
\(=3x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\)
