![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2x+1}{x^2+2}\)
\(\Leftrightarrow Ax^{2\:}+2A=2x+1\)
+) \(A=0\Rightarrow x=-\dfrac{1}{2}\)
+) \(A\ne0\)
\(Ax^2+2A=2x+1\)
\(\Leftrightarrow Ax^{2\:}-2x=1-2A\)
\(\Leftrightarrow x^2-2.\dfrac{x}{A}=\dfrac{1-2A}{A}\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{A}+\dfrac{1}{A^2}=\dfrac{1-2A}{A}+\dfrac{1}{A^2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{A-2A^2+1}{A^2}\)
\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{\left(1-A\right)\left(2A+1\right)}{A^2}\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{A}\right)^2\ge0\left(\forall x,A\ne0\right)\\A^2\ge0\end{matrix}\right.\)
⇒ \(\left(1-A\right)\left(2A+1\right)\ge0\)
⇒ \(-\dfrac{1}{2}\le A\le1\)
Còn lại tụ làm nha
\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2-2+2x+1}{x^2+2}\\ =1-\dfrac{-\left(x-1\right)^2}{x^2+2}\\ Do\left(x-1\right)^2\ge0\Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}\ge0\\ \Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}=0\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}+1\le1\)
\(Dấu"="\Leftrightarrow A=1\\ \Leftrightarrow x-1=0\Rightarrow x=1\\ Vậy.P_{max}=1.khi.x=1\\ A=\dfrac{2x+1}{x^2+2}\rightarrow2A+1=\dfrac{2.\left(2x+1\right)}{x^2+2}+1\\ =\dfrac{4x+2+x^2+2}{x^2+2}=\dfrac{x^2+4x+2}{x^2+2}=\dfrac{\left(x+2\right)^2}{x^2+2}\\ Do\left(x+2\right)^2\ge0\Leftrightarrow\dfrac{\left(x+2\right)^2}{x^2+2}\ge0\)
\(Dấu"="\Leftrightarrow A=\dfrac{1}{2}khi.x=-2\\ \Rightarrow2A+1\ge0\Rightarrow2A\ge-1\Rightarrow A>-\dfrac{1}{2}\\ Vậy.MinA=-\dfrac{1}{2}.khi.x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(A=3\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{4}{9}\right)\)
\(A=\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\)\(\supseteq-\frac{4}{3}\)
Dấu = xr khi x=1/3
Vậy Min A=-4/3 tại x=1/3
\(A=3x^2-2x-1\)
\(=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(=3\left(x^2-2.x.\frac{1}{3}+\frac{1}{9}-\frac{1}{9}-\frac{1}{3}\right)\)
\(=3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\)
Vì \(3\left(x-\frac{1}{3}\right)^2\ge0;\forall x\)
\(\Rightarrow3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\ge0-\frac{4}{3};\forall x\)
Hay \(A\ge\frac{-4}{3};\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=0\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy MIN \(A=\frac{-4}{3}\)\(\Leftrightarrow x=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)
\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)
đẳng thức khi y=-6 thủa mãn đk (*)
Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\dfrac{3x^2+8x+6}{x^2+2x+1}=\dfrac{-2\left(x^2+2x+1\right)+x^2+4x+4}{x^2+2x+1}=-2+\left(\dfrac{x+2}{x+1}\right)^2\ge-2\)
\(S_{min}=-2\) khi \(x=-2\)
Tìm max của C=xy biết 3x+5y=12
Tìm GTNN của: C= x^4 -2x^3+3x^2-4x+2021
Tìm GTNN của D(x)=x^4 -x^2+2x+7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2-3x+1\)
\(=2\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)\)
\(=2\left[\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{16}+\frac{1}{2}\right]\)
\(=2\left[\left(x-\frac{3}{4}\right)^2-\frac{1}{16}\right]\)
\(=2\left(x-\frac{3}{4}\right)^2-\frac{1}{8}\ge\frac{1}{8}\)
Vậy \(Min\left(2x^2-3x+1\right)=\frac{1}{8}\)
Ta có \(2x^2\ge0\)giả sử x=0
\(-3x\ge0\)giả sử x cũng= 0
1>0
\(\Rightarrow2x^2-3x+1\ge1\)
vậy gtnn của:\(2x^2-3x+1\)là 1