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\(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}-1}\div\frac{2}{x-1}+\frac{1}{\sqrt{x}+1}.\)
=\(\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\div\frac{2}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}+1}\)
\(=\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{2+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}\)
\(=\frac{1+x}{\sqrt{x}\times\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{\left(1+x\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1+x}{\sqrt{x}}\)
a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\sqrt{2013-\sqrt{x-1}}=2014-x\)
⇔ \(\left\{{}\begin{matrix}\sqrt{\dfrac{2014-x}{2013+\sqrt{x-1}}}=2014-x\\x\ge1\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}\sqrt{2014-x}.\left(\dfrac{1}{2013+\sqrt{x-1}}-1\right)=0\\x\in\left[1;2014\right]\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}\left[{}\begin{matrix}\dfrac{1}{2013+\sqrt{x-1}}=1\\x=2014\end{matrix}\right.\\x\in\left[1;2014\right]\end{matrix}\right.\)
⇔ x = 2014
Vậy S = {2014}