ĐK: \(m>0,m\ne1\)
a) \(P=\left(\dfrac{\sqrt{m}}{\sqrt{m}-1}-\dfrac{1}{m-\sqrt{m}}\right):\left(\dfrac{1}{\sqrt{m}+1}+\dfrac{2}{m-1}\right)=\left[\dfrac{m}{\sqrt{m}\left(\sqrt{m}-1\right)}-\dfrac{1}{\sqrt{m}\left(\sqrt{m}-1\right)}\right]:\left[\dfrac{\sqrt{m}-1}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}+\dfrac{2}{\left(\sqrt{m}+1\right)\left(\sqrt{m}-1\right)}\right]=\dfrac{m-1}{\sqrt{m}\left(\sqrt{m}-1\right)}:\dfrac{\sqrt{m}-1+2}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}=\dfrac{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}{\sqrt{m}\left(\sqrt{m}-1\right)}:\dfrac{\sqrt{m}+1}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)}=\dfrac{\sqrt{m}+1}{\sqrt{m}}:\dfrac{1}{\sqrt{m}-1}=\dfrac{\left(\sqrt{m}+1\right)\left(\sqrt{m}-1\right)}{\sqrt{m}}=\dfrac{m-1}{\sqrt{m}}\)
b) Thay m=\(3+2\sqrt{2}\) vào P\(\Leftrightarrow P=\dfrac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}=\dfrac{2\sqrt{2}+2}{\sqrt{2+2\sqrt{2}+1}}=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)c) Ta có \(P< 0\Leftrightarrow\dfrac{m-1}{\sqrt{m}}< 0\)(1)
Vì \(\sqrt{m}>0\)
Vậy (1)\(\Leftrightarrow m-1< 0\Leftrightarrow m< 1\)
Kết hợp với ĐK: Vậy m<1 thì P<0