\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{x}{x-1}\right):\left(\sqrt{x}-\dfrac{\sqrt{x}}{\sqrt{x+1}}\right)\)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x+1}}\)
\(M=\dfrac{\sqrt{x^2+x}-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x+1}}{\sqrt{x^2+x}-\sqrt{x}}\)
\(M=\dfrac{\sqrt{x+1}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Để \(M\le0\)
\(\Rightarrow\dfrac{\sqrt{x+1}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\le0\)
mà \(\sqrt{x+1}\ge0\)
\(\Rightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\le0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1\ge0\\\sqrt{x}+1\le0\end{matrix}\right.\)\(\Rightarrow x\ge1\)
\(\left\{{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}+1\ge0\end{matrix}\right.\) \(\Rightarrow x\ge-1\)
Vậy nghiệm của pt : \(x\ge1\) ; \(x\ge-1\)
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P/s: Hm..Câu b mình không chắc chắn lắm /_/
