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DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)
mà \(3x\ge-3\sqrt{5}\)
mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)
min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)
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Xét \(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\) (Đk:\(x\ge2\))
\(2A=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+2\right]+4042\)
\(2A=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)
\(\Leftrightarrow A\le2021\)
\(\Rightarrow Amax=2021\) khi x=3 (tm)Tự đăng câu hỏi xong tự trả lời (T-T)
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\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b) Ta có: \(x\ge0\Rightarrow x+\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{x+\sqrt{x}+1}\le2\)
\(\Rightarrow max=2\) khi \(x=0\)
Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
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$A=2x-\sqrt{x}=2(x-\frac{1}{2}\sqrt{x}+\frac{1}{4^2})-\frac{1}{8}$
$=2(\sqrt{x}-\frac{1}{4})^2-\frac{1}{8}$
$\geq \frac{-1}{8}$
Vậy $A_{\min}=-\frac{1}{8}$. Giá trị này đạt tại $x=\frac{1}{16}$
$B=x+\sqrt{x}$
Vì $x\geq 0$ nên $B\geq 0+\sqrt{0}=0$
Vậy $B_{\min}=0$. Giá trị này đạt tại $x=0$
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\(Đk:x\ge2\)
Đặt \(A=\sqrt{x-2}+2\sqrt{x+1}+2019-x\)
\(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\)
\(=\left[\left(-x+2\right)+2\sqrt{x-2}-1\right]+\left[\left(-x-1\right)+4\sqrt{x+1}-4\right]+4042\)
\(=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+4\right]+4042\)
\(=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)
\(\Rightarrow A\le2021\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{x+1}-2=0\end{matrix}\right.\Leftrightarrow x=3\)
Vậy \(Max\) của biểu thức trên là 2021, đạt tại x=3.