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\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
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\(B=\sqrt{5\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)
\(B_{min}=\frac{\sqrt{3}}{2}\) khi \(x=\frac{1}{2}\)
Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\).Ta có:
\(B\ge\sqrt{5x-4+12-5x}=\sqrt{-\left(4-12\right)}=\sqrt{8}=\sqrt{4}.\sqrt{2}=2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{5x-4}\ge0\\\sqrt{12-5x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}5x\ge4\\5x\le12\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{4}{5}\\x\le\frac{12}{5}\end{cases}\Leftrightarrow\frac{4}{5}\le x\le\frac{12}{5}}\)