1; \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-\left(3\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
2: A<1
=>A-1<0
=>\(\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên và x<>1
nên \(x\in\left\{0;2;3\right\}\)
3) \(mA=\sqrt{x}-2\)
\(\Leftrightarrow\dfrac{2m\sqrt{x}-m}{\sqrt{x}+1}=\sqrt{x}-2\)
\(\Leftrightarrow2m\sqrt{x}-m=x-\sqrt{x}-2\)
\(\Leftrightarrow x-\left(2m+1\right)\sqrt{x}-2+m=0\left(1\right)\)
Để \(\left(1\right)\) có 2 nghiệm phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(2m+1\right)^2-4\left(m-2\right)>0\\S=2m+1>0\\P=m-2>0\\f\left(1\right)=1-2m-1-2+m\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+9>0\left(đúng\right)\\m>-\dfrac{1}{2}\\m>2\\m\ne0\end{matrix}\right.\)
\(\Leftrightarrow m>2\)
Vậy \(m>2\) thỏa mãn đề bài
