hết cứu

\(G=-3x^2-5x+1\\ =-3\left(x^2+2.\dfrac{5}{6}x+\dfrac{25}{36}\right)+\dfrac{37}{12}\\ =\dfrac{37}{12}-3\left(x+\dfrac{5}{6}\right)^2\\ Vì:\left(x+\dfrac{5}{6}\right)^2\ge0\forall x\in R\\ Vậy:G_{max}=\dfrac{37}{12}.khi.x=-\dfrac{5}{6}\)

Lời giải:

$-F=5x^2+4x-3=5(x^2+\frac{4}{5}x+\frac{2^2}{5^2})-\frac{19}{5}$

$=5(x-\frac{2}{5})^2-\frac{19}{5}\geq \frac{-19}{5}$ với mọi $x$

$\Rightarrow F\leq \frac{19}{5}$

Vậy $F_{\max}=\frac{19}{5}$. Giá trị này đạt tại $x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}$

E = - 3\(x^2\) - \(x\) + 2

E = - 3.( \(x^2\) + 2.\(\dfrac{1}{6}\)\(x\) + \(\dfrac{1}{36}\)) + 2

E = -3.(\(x\) + \(\dfrac{1}{6}\))² + \(\dfrac{25}{12}\)

Vì (\(x+\dfrac{1}{6}\))² ≥ 0 ∀ \(x\) ⇒ -3.(\(x+\dfrac{1}{6}\))² ≤ 0 ⇒ -3(\(x+\dfrac{1}{6}\))² + \(\dfrac{25}{12}\) ≤ \(\dfrac{25}{12}\)

E_max = \(\dfrac{25}{12}\) ⇔ \(x\) = - \(\dfrac{1}{6}\) 

Lời giải:

$-E=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

$\Rightarrow E\leq \frac{25}{12}$

Vậy $E_{\max}=\frac{25}{12}$. Giá trị này đạt được khi $x+\frac{1}{6}=0\Leftrightarrow x=\frac{-1}{6}$