a/ \(M=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt[]{x}-1}-1\right)\) (đk: \(x\ge0;x\ne1\))
\(=\left[\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\) :\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}\right)\)
=\(\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{1}{\sqrt{x}-1}\)
=\(\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
vậy M =\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\) với \(x\ge0;x\ne1\)
b/ ta có M=\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\) đk: \(x\ge0;x\ne1\)
để M<3/2 thì:
\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}-\frac{3}{2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\) (vì \(\sqrt{x}+1>0\) do \(x\ge0;x\ne1\))
\(\Leftrightarrow x< 1\) kết hợp với đkxđ
=> \(0\le x< 1\)
vậy \(0\le x< 1\) thì M<3/2
