Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\frac{4\sqrt{5}}{19+2\sqrt{5}}\\ b.B=\ \frac{1}{\sqrt{x} +1} \ với\ x\geqslant 0;x\neq 25\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\frac{4\left(\sqrt{x} +1\right)}{25-x}\\ a.\ Thay\ x=6-2\sqrt{5} =\left(\sqrt{5} -1\right)^{2}\\ \Rightarrow \sqrt{x} =\sqrt{\left(\sqrt{5} -1\right)^{2}} =\sqrt{5} -1\\ vào\ A\ ta\ được:\\ A=\frac{4\left(\sqrt{5} -1+1\right)}{25-\left( 6-2\sqrt{5}\right)} =\frac{4\sqrt{5}}{19+2\sqrt{5}}\\ b.\ B=\left(\frac{15-\sqrt{x}}{x-25} +\frac{2}{\sqrt{x} +5}\right) :\frac{\sqrt{x} +1}{\sqrt{x} -5} ;\ ĐKXĐ:\ x\geqslant 0;x\neq 25\\ =\ \frac{15-\sqrt{x} +2\left(\sqrt{x} -5\right)}{\left(\sqrt{x} +5\right)\left(\sqrt{x} -5\right)} .\frac{\sqrt{x} -5}{\sqrt{x} +1}\\ \equiv \frac{5+\sqrt{x}}{\left(\sqrt{x} +5\right)\left(\sqrt{x} +1\right)} =\frac{1}{\sqrt{x} +1} \end{array}$
