để P>0 thì \(\frac{-\sqrt{x}}{\sqrt{x}-3}\)>0
tiếp theo giải ntn vậy
giúp mk vs ạ
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ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow\sqrt{x-3}=2\sqrt{x^2-9}\)
\(\Leftrightarrow x-3=4\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4\left(x+3\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{4}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x+4}\left(\sqrt{x-4}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-4=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=8\end{matrix}\right.\)
a)\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}}{x-9}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}}{x-9}\)
\(=\frac{3x+9\sqrt{x}+3}{x-9}\)
\(=\)...
\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
Rgọn : P= (\(\frac{x+2}{x\sqrt{x}+1}\) - \(\frac{1}{\sqrt{x}+1}\)) . \(\frac{4\sqrt{x}}{3}\)
= \(\frac{x+2-1\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\). \(\frac{4\sqrt{x}}{3}\)
= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\).\(\frac{4\sqrt{x}}{3}\)
= \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
=>
\(\sqrt{2x+1}-\sqrt{5-x}+x-6=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(1-\sqrt{5-x}\right)+x-4=0\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{5-x}+1}+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{5-x}+1}+1\right)=0\)
\(\Leftrightarrow x=4\)
\(\Leftrightarrow\frac{\sqrt{x}}{3-\sqrt{x}}>0\Leftrightarrow3-\sqrt{x}>0\Rightarrow0< x< 9\)
sai đề r