a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)+\dfrac{3-\sqrt{x}}{x-1}\)
\(=\dfrac{x-\sqrt{x}-x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=-\dfrac{3}{\sqrt{x}+1}\)
c: Để A=-1 thì \(\sqrt{x}+1=3\)
hay x=4
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