ĐKXĐ:...
\(P=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\left(\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}=\frac{2}{\left(\sqrt{x}+1\right)^2}\)
Do \(\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}\ne1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{2}{\left(\sqrt{x}+1\right)^2}< 2\\\frac{2}{\left(\sqrt{x}+1\right)^2}\ne\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\) Để \(M=a\) có nghiệm \(\Leftrightarrow\left\{{}\begin{matrix}0< a< 2\\a\ne\frac{1}{2}\end{matrix}\right.\)
Không rõ đề câu c là gì, bạn viết ko rõ nên khó dịch quá
