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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
\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)
\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Lời giải:
$\frac{3}{2}B=\frac{3\sqrt{x}}{x+\sqrt{x}+1}$
$\Rightarrow 1-\frac{3}{2}B=1-\frac{3\sqrt{x}}{x+\sqrt{x}+1}=\frac{x-2\sqrt{x}+1}{x+\sqrt{x}+1}=\frac{(\sqrt{x}-1)^2}{x+\sqrt{x}+1}\geq 0$ với mọi $x\geq 0$
$\Rightarrow \frac{3}{2}B\leq 1$
$\Rightarrow B\leq \frac{2}{3}$
Vậy $B_{\max}=\frac{2}{3}$ khi $\sqrt{x}-1=0\Leftrightarrow x=1$
\(A=\left(\dfrac{2x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b. Đặt \(B=A-2x\)
\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)
\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)
Cách 1:
Áp dụng BĐT Cô-si:
$x+1\geq 2\sqrt{x}\Rightarrow A=\frac{3\sqrt{x}}{x+1}\leq \frac{3\sqrt{x}}{2\sqrt{x}}=\frac{3}{2}$
Vậy $A_{\max}=\frac{3}{2}$
Giá trị này đạt tại $x=1$
Cách 2:
$\frac{2}{3}A=\frac{2\sqrt{x}}{x+1}$
$\Rightarrow 1-\frac{2}{3}A=1-\frac{2\sqrt{x}}{x+1}=\frac{x-2\sqrt{x}+1}{x+1}=\frac{(\sqrt{x}-1)^2}{x+1}\geq 0$ với mọi $x\geq 0$
$\Rightarrow \frac{2}{3}A\leq 1$
$\Rightarrow A\leq \frac{3}{2}$
Vậy $A_{\max}=\frac{3}{2}$. Giá trị này đạt tại $\sqrt{x}-1=0\Leftrightarrow x=1$
\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)
Ta có \(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}=2\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}=4\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=4-\dfrac{2}{\sqrt{ab}}\)
Khi đó P = \(\dfrac{1}{\sqrt{ab}}\left(4-\dfrac{2}{\sqrt{ab}}\right)=-2\left(\dfrac{1}{\sqrt{ab}}-1\right)^2+2\le2\)
Dấu "=" khi a = b = 1
