ĐKXĐ: \(x\ge 0;x\ne 1\)
a/ \(A=(1+\dfrac{\sqrt x}{x+1}):(\dfrac{1}{\sqrt x-1}-\dfrac{2\sqrt x}{x\sqrt x+\sqrt x-x-1})\\=(\dfrac{x+1}{x+1}+\dfrac{\sqrt x}{x+1}):(\dfrac{1}{\sqrt x-1}-\dfrac{2\sqrt x}{\sqrt x(x+1)-(x+1)})\\=\dfrac{x+\sqrt x+1}{x+1}:(\dfrac{x+1}{(x+1)(\sqrt x-1)}-\dfrac{2\sqrt x}{(x+1)(\sqrt x-1)})\\=\dfrac{x+\sqrt x+1}{x+1}:\dfrac{x+1-2\sqrt x}{(x+1)(\sqrt x-1)}\\=\dfrac{x+\sqrt x+1}{x+1}.\dfrac{(x+1)(\sqrt x-1)}{(\sqrt x-1)^2}\\=(x+\sqrt x+1).\dfrac{\sqrt x-1}{(\sqrt x-1)^2}\\=\dfrac{x+\sqrt x+1}{\sqrt x-1}\)
b/ \(A=7\\→\dfrac{x+\sqrt x+1}{\sqrt x-1}=7\\→x+\sqrt x+1=7(\sqrt x-1)\\↔x+\sqrt x+1=7\sqrt x-7\\↔x-6\sqrt x+8=0\\↔x-2\sqrt x-4\sqrt x+8=0\\↔(x-2\sqrt x)-(4\sqrt x-8)=0\\↔\sqrt x(\sqrt x-2)-4(\sqrt x-2)=0\\↔(\sqrt x-4)(\sqrt x-2)=0\\↔\left[\begin{array}{1}\sqrt x-4=0\\\sqrt x-2=0\end{array}\right.↔\left[\begin{array}{1}\sqrt x=4\\\sqrt x=2\end{array}\right.↔\left[\begin{array}{1}x=16\\x=4\end{array}\right.\)
c/ \(x=2(2+\sqrt 3)\\=4+2\sqrt 3\\=3+2\sqrt 3+1\\=(\sqrt 3+1)^2\)
Thay \(x=(\sqrt 3+1)^2(TM)\) vào biểu thức \(A\)
\(A=\dfrac{(\sqrt 3+1)^2+\sqrt{(\sqrt 3+1)^2}+1}{\sqrt{(\sqrt 3+1)^2}-1}\\=\dfrac{4+2\sqrt 3+|\sqrt 3+1|+1}{|\sqrt 3+1|-1}\\=\dfrac{4+2\sqrt 3+\sqrt 3+1+1}{\sqrt 3+1-1}\\=\dfrac{6+3\sqrt 3}{\sqrt 3}\\=3+2\sqrt 3\)
d/ \(A<1\\→\dfrac{x+\sqrt x+1}{\sqrt x-1}<1\\↔\dfrac{x+\sqrt x+1}{\sqrt x-1}-1<0\\↔\dfrac{x+\sqrt x+1-\sqrt x+1}{\sqrt x-1}<0\\↔\dfrac{x+2}{\sqrt x-1}<0\)
Theo ĐKXĐ: \(x\ge 0→x+2\ge 2\) hay \(x+2>0\)
mà \(\dfrac{x+2}{\sqrt x-1}<0\)
\(→\sqrt x-1<0\\↔\sqrt x<1\\↔x<1\)
Kết hợp ĐKXĐ: \(→0≤x<1\)
Vậy \(0≤x<1\) thì \(A<1\)
