$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\left( 1+\frac{x+\sqrt{x}}{\sqrt{x} +1}\right)\left( 1-\frac{x-\sqrt{x}}{\sqrt{x} -1}\right) +\frac{x-1}{\sqrt{x} +1}\\ DK\ :x\geqslant 0\ ;x\#1\\ A=\left( 1+\frac{\sqrt{x}\left(\sqrt{x} +1\right)}{\sqrt{x} +1}\right)\left( 1-\frac{\sqrt{x}\left(\sqrt{x} -1\right)}{\sqrt{x} -1}\right) +\sqrt{x} -1\\ A=\left( 1+\sqrt{x}\right)\left( 1-\sqrt{x}\right) +\sqrt{x} -1\\ A=1-x+\sqrt{x} -1=\sqrt{x} -x\\ b) \ x=7-4\sqrt{3} \ \\ \rightarrow x=4-2.2\sqrt{3} +3=\left( 2-\sqrt{3}\right)^{2}\\ \rightarrow \sqrt{x} =|2-\sqrt{3} |=2-\sqrt{3}\left( 2 >\sqrt{3}\right) \ \\ A=2-\sqrt{3} -7+4\sqrt{3}\\ A=3\sqrt{3} -5\ \\ c) \ \sqrt{x} -x=-\left( x-2.\frac{1}{2}\sqrt{x} +\frac{1}{4}\right) +\frac{1}{4} \ \\ A=-\left(\sqrt{x} -\frac{1}{2}\right)^{2} +\frac{1}{4}\\ Ta\ có\ :\ -\left(\sqrt{x} -\frac{1}{2}\right)^{2} \leqslant 0\ với\ mọi\ x\ \\ Do\ đó\ :\ -\left(\sqrt{x} -\frac{1}{2}\right)^{2} +\frac{1}{4} \leqslant \frac{1}{4} \ \\ Dấu\ bằng\ xảy\ ra\ khi\ \sqrt{x} -\frac{1}{2} \ =0\ \rightarrow x=\frac{1}{4} \ \\ Vậy\ maxA=\frac{1}{4} \ tại\ x=\frac{1}{4} \end{array}$
