a)
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\\ =\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b)
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\\ \Leftrightarrow3\sqrt{x}-3=\sqrt{x}\\ \Leftrightarrow2\sqrt{x}=3\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\\ \Leftrightarrow x=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\left(nhận\right)\)
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