Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} A=8\\ B=2.\frac{a-1}{a+1} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\left( 2\sqrt{4+\sqrt{\left(\sqrt{5} -1\right)^{2}}}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =\left( 2\sqrt{4+\sqrt{5} -1}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =\left( 2\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =\left( 2\sqrt{2}\sqrt{\frac{3}{2} +\frac{\sqrt{5}}{2}}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =\left( 2\sqrt{2}\sqrt{\left(\frac{\sqrt{5}}{2} +\frac{1}{2}\right)^{2}}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =2\sqrt{2} .\left(\frac{\sqrt{5}}{2} +\frac{1}{2}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =\left(\sqrt{10} +\sqrt{2}\right)\left(\sqrt{10} -\sqrt{2}\right)\\ =10-2=8\\ B=\frac{\left(\sqrt{a} -1\right)^{2} +\left(\sqrt{a} +1\right)^{2}}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\left(\frac{a+1-2}{a+1}\right)^{2}\\ =\frac{2a+2}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\left(\frac{a-1}{a+1}\right)^{2}\\ =\frac{2( a+1)}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\frac{a-1}{a+1} .\frac{a-1}{a+1}\\ =2.\frac{a-1}{a+1} \end{array}$
