Cho A= (4x+5)/(x^2+2x+6). Tim gtln , gtnn 07/08/2021 Bởi Margaret Cho A= (4x+5)/(x^2+2x+6). Tim gtln , gtnn
Đáp án: $Max A=1,MinA=\dfrac{-4}{5}$ Giải thích các bước giải: Ta có: $\begin{split}1-A&=1-\dfrac{4x+5}{x^2+2x+6}\\&=\dfrac{x^2+2x+6-4x-5}{x^2+2x+1+5}\\&=\dfrac{x^2-2x+1}{(x+1)^2+5}\\&=\dfrac{(x-1)^2}{(x+1)^2+5}\\&\ge 0\quad \forall x\end{split}$ $\rightarrow1-A\ge 0$ $\rightarrow A\le 1\rightarrow Max A=1\leftrightarrow x-1=0\rightarrow x=1$ Lại có: $\begin{split}A+\dfrac{4}{5}&=\dfrac{4x+5}{x^2+2x+6}+\dfrac{4}{5}\\&=\dfrac{4(x^2+2x+6)+5(4x+5)}{5(x^2+2x+6}\\&=\dfrac{4x^2+28x+49}{5((x+1)^2+5)}\\&=\dfrac{(2x+7)^2}{5(x+1)^2+25}\\&\ge 0\quad \forall x\end{split}$ $\rightarrow A+\dfrac{4}{5}\ge 0\rightarrow A\ge -\dfrac{4}{5}$ $\rightarrow Min A=\dfrac{-4}{5}\leftrightarrow 2x+7=0\rightarrow x=\dfrac{-7}{2}$ Bình luận
Đáp án:
$Max A=1,MinA=\dfrac{-4}{5}$
Giải thích các bước giải:
Ta có:
$\begin{split}1-A&=1-\dfrac{4x+5}{x^2+2x+6}\\&=\dfrac{x^2+2x+6-4x-5}{x^2+2x+1+5}\\&=\dfrac{x^2-2x+1}{(x+1)^2+5}\\&=\dfrac{(x-1)^2}{(x+1)^2+5}\\&\ge 0\quad \forall x\end{split}$
$\rightarrow1-A\ge 0$
$\rightarrow A\le 1\rightarrow Max A=1\leftrightarrow x-1=0\rightarrow x=1$
Lại có:
$\begin{split}A+\dfrac{4}{5}&=\dfrac{4x+5}{x^2+2x+6}+\dfrac{4}{5}\\&=\dfrac{4(x^2+2x+6)+5(4x+5)}{5(x^2+2x+6}\\&=\dfrac{4x^2+28x+49}{5((x+1)^2+5)}\\&=\dfrac{(2x+7)^2}{5(x+1)^2+25}\\&\ge 0\quad \forall x\end{split}$
$\rightarrow A+\dfrac{4}{5}\ge 0\rightarrow A\ge -\dfrac{4}{5}$
$\rightarrow Min A=\dfrac{-4}{5}\leftrightarrow 2x+7=0\rightarrow x=\dfrac{-7}{2}$