Giải :
a, ĐKXĐ : \(x>1\)
Ta có :
B = \(\dfrac{x+1+\sqrt{x}}{x+1}:\left(\dfrac{1}{\sqrt{x-1}}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
B = \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right).\left(\sqrt{x}-1\right)}\)
B = \(\dfrac{x+1+\sqrt{x}}{x+1}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}\)
B = \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}\)
b, Trong ĐKXĐ ta có :
B > 3 <=> \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}>3\)
<=> \(x+1+\sqrt{x}>3\sqrt{x}-3\)
<=> \(x-2\sqrt{x}+4>0\)
<=> \(\left(\sqrt{x}-1\right)^2+3>0\)
<=> x > 0 ( x \(\in\) R )
Đối chiếu ĐKXĐ ta có :
Với x > 1 thì B > 3
c, Trong ĐKXĐ ta có :
B = 7 <=> \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}=7\)
<=> \(x+1+\sqrt{x}=7\sqrt{x}-7\)
<=> \(x-6\sqrt{x}+8=0\)
<=> \(x-2\sqrt{x}-4\sqrt{x}+8=0\)
<=> \(\left(\sqrt{x}-2\right).\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(TmĐKXĐ\right)\\x=16\left(TmĐKXĐ\right)\end{matrix}\right.\) Vậy với \(x\in\left\{4;16\right\}\)thì B = 7
