a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

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)         

Lời giải:

ĐKXĐ: $x\geq 0$

Với $x\geq 0$ thì $-3\sqrt{x}\leq 0; \sqrt{x}+1>0$. Do đó: $A=\frac{-3\sqrt{x}}{\sqrt{x}+1}\leq 0$

Vậy $A_{\max}=0$. Giá trị này xác định tại $x=0$

\(A=\dfrac{2\sqrt{x}+1-\sqrt{x}}{2\sqrt{x}+1}=1-\dfrac{\sqrt{x}}{2\sqrt{x}+1}\)

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\2\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{2\sqrt{x}+1}\ge0\)

\(\Rightarrow A\le1\)

\(A_{max}=1\) khi \(x=0\)

\(A^2=\left(\sqrt{1-x}+\sqrt{1+x}\right)^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)=4\\ \Leftrightarrow A\le2\\ A_{max}=2\Leftrightarrow1-x=1+x\Leftrightarrow x=0\\ A^2=2+2\sqrt{1-x^2}\ge2\\ \Leftrightarrow A\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow1-x^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(\sqrt{2}\le A\le2\)

Lời giải:
Ta có:
$A^2=x+4+6-x+2\sqrt{(x+4)(6-x)}=10+2\sqrt{(x+4)(6-x)}\geq 10$

$\Rightarrow A\geq \sqrt{10}$ (do $A\geq 0$)

Vậy $A_{\min}=\sqrt{10}$. Giá trị này đạt được khi $(x+4)(6-x)=0\Leftrightarrow x=-4$ hoặc $x=6$

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Áp dụng BĐT Bunhiacopkxy:

$A^2\leq (x+4+6-x)(1+1)=10.2=20$

$\Rightarrow A\leq \sqrt{20}$

Vậy $A_{\max}=\sqrt{20}$

a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

\(=\dfrac{3x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

\(=\dfrac{3x-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

Lời giải:

ĐKXĐ: $x\geq 2$

Áp dụng BĐT AM-GM:

\(\sqrt{x-2}\leq \frac{x-2+2}{2\sqrt{2}}=\frac{x}{2\sqrt{2}}\)

\(\Rightarrow A=\frac{\sqrt{x-2}}{x}\leq \frac{1}{2\sqrt{2}}\)

Vậy $A_{\max}=\frac{1}{2\sqrt{2}}$. Giá trị này đạt tại $x-2=2\Leftrightarrow x=4$