a) Để biểu thức M có nghĩa thì \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) \(M=\frac{2}{\sqrt{x}-1}+\frac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}=\frac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{2x+2\sqrt{x}+2+2x-2+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)c) Ta có \(M=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\Leftrightarrow Mx+M\sqrt{x}+M-5\sqrt{x}+3=0\Leftrightarrow Mx+\left(M-5\right)\sqrt{x}+\left(M+3\right)=0\)Để phương trình có nghiệm( hay có giá trị x) thì \(\left(M-5\right)^2-4.M.\left(M+3\right)\ge0\Leftrightarrow M^2-10M+25-4M^2-12M\ge0\Leftrightarrow3M^2+22M-25\le0\Leftrightarrow\left(M-1\right)\left(3M+25\right)\le0\Leftrightarrow\)\(-\frac{25}{3}\le M\le1\)
Vậy M có GTLN khi \(\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}=1\Leftrightarrow x+\sqrt{x}+1=5\sqrt{x}-3\Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow x=4\)
Vậy để biểu thức M có GTLN là 1 thì x=4
