\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b.
\(mA=\sqrt{x}-2\Rightarrow\dfrac{m\left(2\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-2\)
\(\Rightarrow2m\sqrt{x}+m=\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow x-\left(2m+1\right)\sqrt{x}-2-m=0\)
Đặt \(\sqrt{x}=t\ge0\)
\(\Rightarrow t^2-\left(2m+1\right)t-m-2=0\) (1)
Pt đã cho có 2 nghiệm pb khi (1) có 2 nghiệm dương pb
\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(2m+1\right)^2+4\left(m+2\right)>0\\t_1+t_2=2m+1>0\\t_1t_2=-m-2>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4\left(m+1\right)^2+1>0\\m>-\dfrac{1}{2}\\m< -2\end{matrix}\right.\)
\(\Rightarrow\) Không tồn tại m thỏa mãn
c.
\(A=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=1+\dfrac{\sqrt{x}}{\sqrt{x}+1}\ge1\) do \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\ge0\) với mọi \(x\ge0\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)-1}{\sqrt{x}+1}=2-\dfrac{1}{\sqrt{x}+1}< 2\) do \(\dfrac{1}{\sqrt{x}+1}>0\)
\(\Rightarrow1\le A< 2\)
Mà A nguyên \(\Rightarrow A=1\)
\(\Rightarrow x=0\)
